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<span class="text_page_counter">Trang 1</span><div class="page_container" data-page="1">
Chu de 1. Ham
1.1. Tinh dan <small>di~u </small>cua ham so ... 8
1.2. Cl;l'c tri ham so ... 13
1.3. Gia tri I6n nhat, gia tri nho nhat ... 19
1.4. Duemg <small>ti~m c~n </small>cua do thi ham so ... 20
1.5. Do thi ham so chua dau gia tri <small>tuy~t </small>doi ... 21
1.6. Tiep tuyen ... 22
1.7. Tuang giao do tht ... 23
Chu de 2. Liiy thira, mii va logarit ... 28
2.1. Ham so lily thua ... 28
2.2. Ham so mil ... 28
2.3,. Logarit va ham so logarit ... 29
2.4. Phuang trlnh mil ... 30
2.5. Phuang trlnh logarit ... 32
2.6. Bat phuO'Ilg trlnh mil ... 33
2.7. Bat phuang trlnh logarit ... 34
2.8. Bai to an lai suat ngan hang ... 36
Chu de 3. Nguyen ham- tfch phan- Un.g d\lng tich phan ... 38
3.1. Bang nguyen ham cua <i><small>m<)t </small></i>so ham thuang <small>g~p </small>... , ... 38
3.2. Cac phuO'Ilg phap tinh nguyen ham ... 38
3.3. Cac phuO'Ilg phap tinh tich phan ... .41
3.4. Tich phan cac ham so so cap co ban ... 43
3.5. Ung dvng tich phan ... 45
3.6. M<)t so phuang phap tinh tich phan ham an ... 50
3.7. M<)t so thu <small>thu~t </small>su dvng casio ... 51
Chu de 4. Sd phuc ... 53
4.1. Kien thuc co ban ... 53
4.2. Cac phep to an co ban so phuc ... 53
4 3 T. . c;tp Q'P 1em 1eu 1en sop uc ... . <small>A </small> h d' <small>A? </small> b' <small>A? </small> d' <small>J:: </small> <i><small>AI </small></i> h <i><small>r </small></i> 55 4.4. Phuang trlnh <small>b~c </small>hai v6i <small>h~ </small>so thl;l'c ... 59
4.5. M<)t so bai toan lien quan den min- max so phuc ... 60
4.6. M<)t so thu <small>thu~t </small>su dvng may tinh (casio) ... 61
Chu s. da 5.1. Khai <small>ni~m </small>ve hlnh da <small>di~n </small>va khoi da <small>di~n </small>... 63
5.2. The tich khoi da <small>di~n </small>... 64
5.3 M<)t so cong thuc tinh nhanh the' tich khoi chop thuemg gi{tp ... 68
5.4. M<)t so cong thuc tinh nhanh ve ill <small>di~n </small>... 72
5.5. M<)t so cong thuc tinh nhanh ti so the tich ... 73
6.3. M.;it cau ... 77
6.4. M<)t so <small>d~ng </small>to an va cong thuc giai nhanh m.;it non ... 78
<small>d~ng </small>Chu de 1. Ham
1.1. Tinh dan <small>di~u </small>cua ham so ... 8
1.2. Cl;l'c tri ham so ... 13
1.3. Gia tri I6n nhat, gia tri nho nhat ... 19
1.4. Duemg <small>ti~m c~n </small>cua do thi ham so ... 20
1.5. Do thi ham so chua dau gia tri <small>tuy~t </small>doi ... 21
1.6. Tiep tuyen ... 22
1.7. Tuang giao do tht ... 23
Chu de 2. Liiy thira, mii va logarit ... 28
2.1. Ham so lily thua ... 28
2.2. Ham so mil ... 28
2.3,. Logarit va ham so logarit ... 29
2.4. Phuang trlnh mil ... 30
2.5. Phuang trlnh logarit ... 32
2.6. Bat phuO'Ilg trlnh mil ... 33
2.7. Bat phuang trlnh logarit ... 34
2.8. Bai to an lai suat ngan hang ... 36
Chu de 3. Nguyen ham- tfch phan- Un.g d\lng tich phan ... 38
3.1. Bang nguyen ham cua <i><small>m<)t </small></i>so ham thuang <small>g~p </small>... , ... 38
3.2. Cac phuO'Ilg phap tinh nguyen ham ... 38
3.3. Cac phuO'Ilg phap tinh tich phan ... .41
3.4. Tich phan cac ham so so cap co ban ... 43
3.5. Ung dvng tich phan ... 45
3.6. M<)t so phuang phap tinh tich phan ham an ... 50
3.7. M<)t so thu <small>thu~t </small>su dvng casio ... 51
Chu de 4. Sd phuc ... 53
4.1. Kien thuc co ban ... 53
4.2. Cac phep to an co ban so phuc ... 53
4 3 T. . c;tp Q'P 1em 1eu 1en sop uc ... . <small>A </small> h d' <small>A? </small> b' <small>A? </small> d' <small>J:: </small> <i><small>AI </small></i> h <i><small>r </small></i> 55 4.4. Phuang trlnh <small>b~c </small>hai v6i <small>h~ </small>so thl;l'c ... 59
4.5. M<)t so bai toan lien quan den min- max so phuc ... 60
4.6. M<)t so thu <small>thu~t </small>su dvng may tinh (casio) ... 61
Chu s. da 5.1. Khai <small>ni~m </small>ve hlnh da <small>di~n </small>va khoi da <small>di~n </small>... 63
5.2. The tich khoi da <small>di~n </small>... 64
5.3 M<)t so cong thuc tinh nhanh the' tich khoi chop thuemg gi{tp ... 68
5.4. M<)t so cong thuc tinh nhanh ve ill <small>di~n </small>... 72
5.5. M<)t so cong thuc tinh nhanh ti so the tich ... 73
6.3. M.;it cau ... 77
6.4. M<)t so <small>d~ng </small>to an va cong thuc giai nhanh m.;it non ... 78 <small>d~ng </small>
</div><span class="text_page_counter">Trang 4</span><div class="page_container" data-page="4">6.6. M9t so <small>d~mg </small>toan va cong thuc giaitoan mij.t cau ... 92
6.7. Tong hqp cac cong thuc dij.c <small>bi~t </small>ve khoi tron xoay ... 100
Chit de 7. Phuong phap tQa d(} trong khong gian ... 101
7.1. <small>H~ </small>t9a d9 trong khong gian ... 101
7.2. Phuotlg trlnh mij.t cau ... 102
7.3. Mij.t ph~ng ... 103
' h.:!, 7.4. Duotlg t ang ... 104
7.5. Cac bai toan tr9ng tam ve t9a d9 ... 106
7.6. Cac bai toan tr9ng tam ve mij.t ph~ng ... 108
7.7. Cac bai toan tr9ng tam ve duemg th~ng ... 115
7.8. Cac bai toan tr9ng tam ve mij.t cau ... 126
Chit de 8. <i>T6 </i>hqp- xac sua't, ca'p so, g6c va khoang each ... 132
8.1. Hoan vi - chinh hqp - to hqp ... 132
8.2. Nhi thuc newton ... · ... 133
8.3. Xac suat ... 135
8.4. Cap so c9ng, cap so nhan ... 137
8.5. Cach xach dinh g6c trong khong gian ... 138
8.6. Cach xach dinh khoang each trong khong gian ... 141
A- 300 cau hoi trQng tam ... 146
1. Ham so va cac bai toan lien quan ... 146
2. Ham so lily thua, ham so mil va ham so logarit ... 157
3. Nguyen ham, tich phan va U>ng dvng ... 161
4. so phuc ... 166
5. Hlnh h9c khong gian ... 169
6. Mij.t n6n, mij.t trv, mij.t c'i\l.u ... 172
7. Phuotlg phap t9a d9 khong gian ... 174
8. To hqp- xac suat, cap so c9ng- cap so nhan ... 179
B- Dap an va huong d~n giai chi tie't ... 182
1. Ham so va cac bai toan lien quan ... 182
2. Ham so lily thua, ham so mil va ham so logarit ... 195
3. Nguyen ham, tich phan va U>ng dvng ... 202
4. so phuc ... 209
5. Hlnh h9c khong gian ... 215
6. Mij.t n6n, mij.t trv, mij.t cau ... 224
7. Phuotlg phap t9a d9 trong khong gian ... 228
8. To hqp- xac suat, cap so c9ng- cap so nhan ... 241
A - B(} cau hoi hay -<small>1~ </small>- kh6 hQc tu duy 10 diem ... 246
B - Dap an va <small>d~n </small>giai chi tie't ... 262
LAM 1. Chie'n thu~t t6ng on, luy~n de tang 2-3 diem giai do~n nuoc rut ... 313
2. Bi kip phan b6 thai gian lam de hi~u qua ...
6.6. M9t so <small>d~mg </small>toan va cong thuc giaitoan mij.t cau ... 92
6.7. Tong hqp cac cong thuc dij.c <small>bi~t </small>ve khoi tron xoay ... 100
Chit de 7. Phuong phap tQa d(} trong khong gian ... 101
7.1. <small>H~ </small>t9a d9 trong khong gian ... 101
7.2. Phuotlg trlnh mij.t cau ... 102
7.3. Mij.t ph~ng ... 103
' h.:!, 7.4. Duotlg t ang ... 104
7.5. Cac bai toan tr9ng tam ve t9a d9 ... 106
7.6. Cac bai toan tr9ng tam ve mij.t ph~ng ... 108
7.7. Cac bai toan tr9ng tam ve duemg th~ng ... 115
7.8. Cac bai toan tr9ng tam ve mij.t cau ... 126
Chit de 8. <i>T6 </i>hqp- xac sua't, ca'p so, g6c va khoang each ... 132
8.1. Hoan vi - chinh hqp - to hqp ... 132
8.2. Nhi thuc newton ... · ... 133
8.3. Xac suat ... 135
8.4. Cap so c9ng, cap so nhan ... 137
8.5. Cach xach dinh g6c trong khong gian ... 138
8.6. Cach xach dinh khoang each trong khong gian ... 141
A-300 cau hoi trQng tam ... 146
1. Ham so va cac bai toan lien quan ... 146
2. Ham so lily thua, ham so mil va ham so logarit ... 157
3. Nguyen ham, tich phan va U>ng dvng ... 161
4. so phuc ... 166
5. Hlnh h9c khong gian ... 169
6. Mij.t n6n, mij.t trv, mij.t c'i\l.u ... 172
7. Phuotlg phap t9a d9 khong gian ... 174
8. To hqp- xac suat, cap so c9ng- cap so nhan ... 179
B- Dap an va huong d~n giai chi tie't ... 182
1. Ham so va cac bai toan lien quan ... 182
2. Ham so lily thua, ham so mil va ham so logarit ... 195
3. Nguyen ham, tich phan va U>ng dvng ... 202
4. so phuc ... 209
5. Hlnh h9c khong gian ... 215
6. Mij.t n6n, mij.t trv, mij.t cau ... 224
7. Phuotlg phap t9a d9 trong khong gian ... 228
8. To hqp- xac suat, cap so c9ng- cap so nhan ... 241
A - B(} cau hoi hay -<small>1~ </small>- kh6 hQc tu duy 10 diem ... 246
B - Dap an va <small>d~n </small>giai chi tie't ... 262
LAM 1. Chie'n thu~t t6ng on, luy~n de tang 2-3 diem giai do~n nuoc rut ... 313 2. Bi kip phan b6 thai gian lam de hi~u
</div><span class="text_page_counter">Trang 5</span><div class="page_container" data-page="5">Ky thi THPT Quae gia dang den rat gan, m()t buoc nga~t Ian doi voi mbi si
L3.m the naa de' but pha
Co 5 yeu to quan trQng quyet dinh tai die'm so cua hQC sinh trang giai da9-n nuac rut nay chinh I
1. Kien thuc
2. Chien <small>thu~t </small>hQc tang diem 3. Toe d9 lam de, phim x9- de 4. Tam ly phong thi vfmg vang 5. Sv cham chi, kien td
On sai each, Iuoi bieng, chenh mimg se khien lt;I'C hQC cua b9-n mai d~m chan t9-i chb. Giai phap tot nhat cha cac b9-n can ngay bay gio chinh la b9 sach <i><small>I I </small></i>CAP TOC 789+" tong on thi THPT QG voi 4 mon quan trQng: Taan, Li, Hoa, Anh se giup b9-n:
• <small>T~p </small>trung on dung, du, trung phan kien thuc <small>t~p </small>trung nhat, tranh on Ian man, sa da khong dung mvc tieu diem so
• Cung cap bQ cau hoi ch~t IQc nhat vm nhfrng d9-ng bai muc d@, kho, Ct;I'C kho co kha nang caa xuat <small>hi~n </small>trang de thi .
• Tang cu&ng phan x9- de, thu n9-p nhieu phuang phap giai t~t, nham nhanh, daan y giup tang <small>3 </small>toe d9 lam de
• B6 tUi nhi'eu m~a, tips tranh b§.y hay la kinh nghi~m col-0-2
• Va <small>d~c bi~t </small>khong the thieu nhfmg I&i khuyen hfru ich de' vaa phong thi co tam ly thaai mainhat
Day cling Ia b9 sach <small>d~c bi~t </small>co phan chia cac cap de) kien thuc thea rung mvc tieu diem so 7+, 8+, 9+ phu hQp vai mang muon diem thi clia mbi b9-n, giup si
Voi kinh <small>nghi~m </small>nhi'eu nam trang qua tdnh giang d9-y Taan pho thong cling kinh <small>nghi~m </small>giai, phan tich de thi THPT Quae gia, sv nh9-y ben trang phan daan xu huang ra de ket hQ'p n9i dung bam sat rna <small>tr~n </small>de thi chuan se giup cac b9-ll on <small>t~p </small>m9t each <small>hi~u </small>qua nhat.
Trang qua trinh bien sa9-TI ch~c ch~n khong tranh het du9c nhfrng sai sot, nham l§.n, mang b9-n dQC thong cam va chia se, dong gop nhfmg th~c m~c de cuon sach haan thi~n han trang nhfrng fan t.ii ban sau.
Le Due <small>Thi~u </small>(Chu bien)
Ky thi THPT Quae gia dang den rat gan, m()t buoc nga~t Ian doi voi mbi si
L3.m the naa de' but pha
Co 5 yeu to quan trQng quyet dinh tai die'm so cua hQC sinh trang giai da9-n nuac rut nay chinh I
1. Kien thuc
2. Chien <small>thu~t </small>hQc tang diem 3. Toe d9 lam de, phim x9- de 4. Tam ly phong thi vfmg vang 5. Sv cham chi, kien td
On sai each, Iuoi bieng, chenh mimg se khien lt;I'C hQC cua b9-n mai d~m chan t9-i chb. Giai phap tot nhat cha cac b9-n can ngay bay gio chinh la b9 sach <i><small>I I </small></i>CAP TOC 789+" tong on thi THPT QG voi 4 mon quan trQng: Taan, Li, Hoa, Anh se giup b9-n:
• <small>T~p </small>trung on dung, du, trung phan kien thuc <small>t~p </small>trung nhat, tranh on Ian man, sa da khong dung mvc tieu diem so
• Cung cap bQ cau hoi ch~t IQc nhat vm nhfrng d9-ng bai muc d@, kho, Ct;I'C kho co kha nang caa xuat <small>hi~n </small>trang de thi .
• Tang cu&ng phan x9- de, thu n9-p nhieu phuang phap giai t~t, nham nhanh, daan y giup tang <small>3 </small>toe d9 lam de
• B6 tUi nhi'eu m~a, tips tranh b§.y hay la kinh nghi~m col-0-2
• Va <small>d~c bi~t </small>khong the thieu nhfmg I&i khuyen hfru ich de' vaa phong thi co tam ly thaai mainhat
Day cling Ia b9 sach <small>d~c bi~t </small>co phan chia cac cap de) kien thuc thea rung mvc tieu diem so 7+, 8+, 9+ phu hQp vai mang muon diem thi clia mbi b9-n, giup si
Voi kinh <small>nghi~m </small>nhi'eu nam trang qua tdnh giang d9-y Taan pho thong cling kinh <small>nghi~m </small>giai, phan tich de thi THPT Quae gia, sv nh9-y ben trang phan daan xu huang ra de ket hQ'p n9i dung bam sat rna <small>tr~n </small>de thi chuan se giup cac b9-ll on <small>t~p </small>m9t each <small>hi~u </small>qua nhat.
Trang qua trinh bien sa9-TI ch~c ch~n khong tranh het du9c nhfrng sai sot, nham l§.n, mang b9-n dQC thong cam va chia se, dong gop nhfmg th~c m~c de cuon sach haan thi~n han trang nhfrng fan t.ii ban sau.
Le Due <small>Thi~u </small>(Chu bien)
</div><span class="text_page_counter">Trang 8</span><div class="page_container" data-page="8">Cho ham so y
+ Neu f <small>I </small>
+ Neu f <small>I </small>
D "'· <small>01 </small>va1 a m y = - -,. h' ax+b ( x <small>:;t: - -</small>dJ kh' ' tinhd 1xet an d'" <small>1~u </small>t hl f'( ) 0 x = kh" ongxayra. en ta ' N."
c>O
Khi g ( x) = 0 co 2 nghi~m phan bi~t x<sub>1 ; </sub>x<sub>2 </sub>thl a.g ( x) < 0 Vx <small>E ( </small>x<sub>1; X 2 ) </sub>
{a<O 11::;0
Tim m de' ham so y = f ( x) dong bien, nghich bien tren khoang (a;
Su dvng each xet dau tam thuc <small>b~c </small>2 (dung cho khi khong co <small>l~p </small>duqc m) Vi~c xet dau f'
Dua ve bai toan min- max (dung cho khi co <small>l~p </small>duqc m)
tht,rc). Co bao nhieu gia tri nguyen cua m de ham soda cho dong bien tren khoang ( 0; +oo)?
Cho ham so y
+ Neu f <small>I </small>
+ Neu f <small>I </small>
D "'· <small>01 </small>va1 a m y = - -,. h' ax+b ( x <small>:;t: - -</small>dJ kh' ' tinhd 1xet an d'" <small>1~u </small>t hl f'( ) 0 x = kh" ongxayra. en ta ' N."
c>O
Khi g ( x) = 0 co 2 nghi~m phan bi~t x<sub>1 ; </sub>x<sub>2 </sub>thl a.g ( x) < 0 Vx <small>E ( </small>x<sub>1; X 2 ) </sub>
{a<O 11::;0
Tim m de' ham so y = f ( x) dong bien, nghich bien tren khoang (a;
Su dvng each xet dau tam thuc <small>b~c </small>2 (dung cho khi khong co <small>l~p </small>duqc m) Vi~c xet dau f'
Dua ve bai toan min- max (dung cho khi co <small>l~p </small>duqc m)
</div><span class="text_page_counter">Trang 9</span><div class="page_container" data-page="9">Di'eu ki~n xac dinh: <small>X </small>
(x-m)
Ham soda cho dong bien tren khoang ( o; +oo)
<=> 4 m<small>2 </small>
f'(x)= - <sub>2 </sub>>0 VxE(O;+oo) <sup><=> </sup>{ <sup><=> </sup> <sup><=> -2 < m s 0. </sup>
4-m<small>2 </small>
>0 -2<m<2 (x-m)
cua m th6a man cau bai toan. Biet t?p hqp tat ca cac gia tri tht,l'C cua tham so m de? ham so
y
T?p xac dinh D = 1R . Ta c6 y' = 6x<small>2 </small>
Ham so luon dong bien tren 1R => y' <small>~ </small>o, v x E 1R
<small><=>~so </small>
<=> (2m+1r -4m(m+1) so <=> 1 s 0 (L)
Phuong tdnh y' <small>= </small>0 c6 hai <small>nghi~m </small>phan <small>bi~t </small>th6a man x<sub>1 </sub>< x<sub>2 </sub>s 2 <=> x<sub>1 </sub><small>-</small> 2 < x<sub>2 </sub><small>-</small> 2 s 0
+ 14mx+5 s 0, Vx ~ 1 <=> g(x) <small>= </small>---~ m (1) +14x
C6 g'(x) = <sub>2 </sub>> 0, Vx E [ 1; +oo), suy ra ming(x) = g(1) = --.
Di'eu ki~n xac dinh: <small>X </small>
(x-m)
Ham soda cho dong bien tren khoang ( o; +oo)
<=> 4 m<small>2 </small>
f'(x)= - <sub>2 </sub>>0 VxE(O;+oo) <sup><=> </sup>{ <sup><=> </sup> <sup><=> -2 < m s 0. </sup>
4-m<small>2 </small>
>0 -2<m<2 (x-m)
cua m th6a man cau bai toan. Biet t?p hqp tat ca cac gia tri tht,l'C cua tham so m de? ham so
y
T?p xac dinh D = 1R . Ta c6 y' = 6x<small>2 </small>
Ham so luon dong bien tren 1R => y' <small>~ </small>o, v x E 1R
<small><=>~so </small>
<=> (2m+1r -4m(m+1) so <=> 1 s 0 (L)
Phuong tdnh y' <small>= </small>0 c6 hai <small>nghi~m </small>phan <small>bi~t </small>th6a man x<sub>1 </sub>< x<sub>2 </sub>s 2 <=> x<sub>1 </sub><small>-</small> 2 < x<sub>2 </sub><small>-</small> 2 s 0
+ 14mx+5 s 0, Vx ~ 1 <=> g(x) <small>= </small>---~ m (1) +14x
C6 g'(x) = <sub>2 </sub>> 0, Vx E [ 1; +oo), suy ra ming(x) = g(1) = --.
</div><span class="text_page_counter">Trang 10</span><div class="page_container" data-page="10">Xet Sl,f tuang giao cua do thi ham so so y
Xet Sl,f tuang giao cua do thi ham so so y
Dl,fa vao do thj ta co g'
Cho ham so y = £ (X) co dl;lO ham tren ~ thoa man f ( 2) = f ( -2) = 0 Va dO thi CUa ham SO Y = f <small>1 </small>
(X) CO d1;1ng nhU hlnh ben. Ham so y = ( f (X)
( -1;
Ta co £I (x) = 0 <=> x = 1; x = ±2; £ ( 2) = £ ( -2) = 0 . Ta co bang bien thien:
=> f(x) < O;'v'x
Taco g'(x) = 2f'(x).f(x). Xet g'(x) < 0 <=> f'(x).f(x) < 0 <=> {f'((x)) > <sup>0 </sup><=> [x < -<sup>2 </sup> . £ x <0 1<x<2 Suy ra ham so g (x) nghich bien tren cac khoang ( -oo; -2) <i><small>I </small></i> <small>( </small>1; 2).
Cho ham so y = £ ( x). Do thi ham so y = f' ( x) nhu hinh ben. Ham so g (X) = £ (13- xl) dong bien tren khoang nao du6i day?
( -oo;1). ( 2;4 ).
Dung MODE 7 <small>nh~p </small><i>ham; quan sat bang gia tri ham so; khoang nElO bang gia tri luon </i>
tang thl ham so dong bien tren khoang do; khoang nao bang gia tri luon giam thl ham so nghich
Cho ham so y = £ (X) co dl;lO ham tren ~ thoa man f ( 2) = f ( -2) = 0 Va dO thi CUa ham SO Y = f <small>1 </small>
(X) CO d1;1ng nhU hlnh ben. Ham so y = ( f (X)
( -1;
Ta co £I (x) = 0 <=> x = 1; x = ±2; £ ( 2) = £ ( -2) = 0 . Ta co bang bien thien:
=> f(x) < O;'v'x
Taco g'(x) = 2f'(x).f(x). Xet g'(x) < 0 <=> f'(x).f(x) < 0 <=> {f'((x)) > <sup>0 </sup><=> [x < -<sup>2 </sup> . £ x <0 1<x<2 Suy ra ham so g (x) nghich bien tren cac khoang ( -oo; -2) <i><small>I </small></i> <small>( </small>1; 2).
Cho ham so y = £ ( x). Do thi ham so y = f' ( x) nhu hinh ben. Ham so g (X) = £ (13- xl) dong bien tren khoang nao du6i day?
( -oo;1). ( 2;4 ).
Dung MODE 7 <small>nh~p </small><i>ham; quan sat bang gia tri ham so; khoang nElO bang gia tri luon </i>
tang thl ham so dong bien tren khoang do; khoang nao bang gia tri luon giam thl ham so nghich
</div><span class="text_page_counter">Trang 12</span><div class="page_container" data-page="12">Su dl,lng d9-0 ham t9-i 1 die'm neu <small>Xo) </small> 0; <small>Xo E </small>(a; b) thl ham so khong dong bien tren khming (a; b); neu f' ( <small>X</small><sub>0 ) </sub>> 0; <small>X</small><sub>0 </sub><small>E </small>(a; b) thl. ham SO khong nghich bien tren khm\ng (a; b).
Ham so y =
dong bien tren khoimg nao sau day? ( 1; +
<b>chQn START: -5; END: 4: STEP: 0.5 </b>
thi
-0 5
Dung CALC de thu cac gia tri trong khoang can xet
Bam CALC gia tri X ta b6 qua; bam dau "=" va gan A gia tri 1,1 ta thay ket qua am tuc mQi khoang chua <small>X </small>= 1,1 thl khong dong bien. <small>V~y </small>lo9-i dap an A va dap
Con dap an C va dap an D thl tiep tl,lc tinh d9-o ham t9-i 1 diem thu()c ( -oo; 1) rna khong thu()c ( -2;1) vi d1,1 nhu x <small>= </small>-5 ta duqc ket qua la Math ERRO. V~y
uoc tinh Tinh d9-o ham t9-i die'm x=1,1
Tinh d9-o ham t9-i die'm x=-5
c.:snce J: Giot.o
T, 1111 tat ca cac gm <sup>" ' ' , , </sup> <sup>., </sup><small>tq </small><sup>. h </sup>t vc cua <sup>, </sup> <small>t </small><sup>h </sup>am so <sup>,.., </sup>m <sup>dA'h' </sup>e am so y <sup>,.., </sup> <small>= </small><sup>cosx-2 </sup> ng .c <sup>hihb'"' </sup>1en tren <sup>A </sup>cosx-m
D :;;: h"" <sub>e t ay va1 m = t </sub>,. 2 hl <sub>y = </sub>cosx- 2 <sub>= </sub>1 A h' <sub>nen am so </sub>""khA <sub>ong </sub>d an d'A <small>1~u </small>tren A kh ' oang
<b>chQn START: -5; END: 4: STEP: 0.5 </b>
thi
-0 5
Dung CALC de thu cac gia tri trong khoang can xet
Bam CALC gia tri X ta b6 qua; bam dau "=" va gan A gia tri 1,1 ta thay ket qua am tuc mQi khoang chua <small>X </small>= 1,1 thl khong dong bien. <small>V~y </small>lo9-i dap an A va dap
Con dap an C va dap an D thl tiep tl,lc tinh d9-o ham t9-i 1 diem thu()c ( -oo; 1) rna khong thu()c ( -2;1) vi d1,1 nhu x <small>= </small>-5 ta duqc ket qua la Math ERRO. V~y
uoc tinh Tinh d9-o ham t9-i die'm x=1,1
Tinh d9-o ham t9-i die'm x=-5
c.:snce J: Giot.o
T, 1111 tat ca cac gm <sup>" ' ' , , </sup> <sup>., </sup><small>tq </small><sup>. h </sup>t vc cua <sup>, </sup> <small>t </small><sup>h </sup>am so <sup>,.., </sup>m <sup>dA'h' </sup>e am so y <sup>,.., </sup> <small>= </small><sup>cosx-2 </sup> ng .c <sup>hihb'"' </sup>1en tren <sup>A </sup>cosx-m
D :;;: h"" <sub>e t ay va1 m = t </sub>,. 2 hl <sub>y = </sub>cosx- 2 <sub>= </sub>1 A h' <sub>nen am so </sub>""khA <sub>ong </sub>d an d'A <small>1~u </small>tren A kh ' oang
Thay cac gia <i>tri cua F </i>
<i>+ Buac 1: </i>Tim tap xac dinh. Tfnh f'(x).
<i>+ Buac 2: </i>Tim cac diem xi rna tc;ti d6 f'
<i>+ Buac 3: </i>Lap bang bien thien
<i>+ Buac 4: </i>Neu f' ( x) doi dau khi di qua xi thi ham
<i>+ Buac 1: </i>Tim tap xac dinh. Tim f'(x).
<i>+ Buac 2: </i>Tim cac nghi~m xi (i
x<sub>1 </sub>< < 0 P > 0
£) x<sub>1 </sub>< a < x<sub>2 </sub> <=> ( x<sub>1 </sub><small>-</small> a) ( x<sub>2 -</sub> a) < 0 <=> x<sub>1 </sub>.x<sub>2 -</sub> a ( x<sub>1 </sub>+ x<sub>2 ) </sub>+ a <small>2 </small>< 0
( x<small>1 </small>-a) ( x<small>2 </small> a) > 0 {x<small>1 </small>.x<small>2 -</small> a ( x<small>1 </small>+ x<small>2 ) </small>+ a <sup>2 </sup>> 0 g) < < a <=> <=>
x<sub>1 </sub>+ x<sub>2 </sub>< 2a x<sub>1 </sub>+ x<sub>2 </sub>< 2a <small>,__,... {( X 1 -</small>
> 0 h) <small>a </small>< <small>x</small><sub>1 </sub>< <i><small>'r-7 'r-7 </small></i>
x + x > 2a x + x > 2a
Thay cac gia <i>tri cua F </i>
<i>+ Buac 1: </i>Tim tap xac dinh. Tfnh f'(x).
<i>+ Buac 2: </i>Tim cac diem xi rna tc;ti d6 f'
<i>+ Buac 3: </i>Lap bang bien thien
<i>+ Buac 4: </i>Neu f' ( x) doi dau khi di qua xi thi ham
<i>+ Buac 1: </i>Tim tap xac dinh. Tim f'(x).
<i>+ Buac 2: </i>Tim cac nghi~m xi (i
x<sub>1 </sub>< < 0 P > 0
£) x<sub>1 </sub>< a < x<sub>2 </sub> <=> ( x<sub>1 </sub><small>-</small> a) ( x<sub>2 -</sub> a) < 0 <=> x<sub>1 </sub>.x<sub>2 -</sub> a ( x<sub>1 </sub>+ x<sub>2 ) </sub>+ a <small>2 </small>< 0
( x<small>1 </small>-a) ( x<small>2 </small> a) > 0 {x<small>1 </small>.x<small>2 -</small> a ( x<small>1 </small>+ x<small>2 ) </small>+ a <sup>2 </sup>> 0 g) < < a <=> <=>
x<sub>1 </sub>+ x<sub>2 </sub>< 2a x<sub>1 </sub>+ x<sub>2 </sub>< 2a <small>,__,... {( X 1 -</small>
> 0 h) <small>a </small>< <small>x</small><sub>1 </sub>< <i><small>'r-7 'r-7 </small></i>
x + x > 2a x + x > 2a
</div><span class="text_page_counter">Trang 14</span><div class="page_container" data-page="14">+ A, B n~m cimg ve 1 phia doi v6i trt.tc Oy <=> y' = 0 co hai nghi~m phan bi~t cling dau + A, B n~m cling ve 2 phia doi v6i tn,1.c Oy <=> y' = co hai nghi~m trcli dau
+A, B n~m cling ve 1 phia doi v6i trt,lC Ox <small>¢:> </small>y' = 0 co hai nghi~m phan bi~t va y <small>CD </small>.y <small>cr </small>> 0
Yco+Ycr>O
+A, B
A B
Yco·Ycr <0
(ap dvng khi khong nham duqc nghi~m va viet duqc phuang trlnh duemg th~ng di qua hai diem Cl,fC tri cua do thi ham so)
+ A, B n~m ve 2 phia doi v6i tn.tc Ox
<=> do thi c~t trvc Ox t;;1.i 3 diem phan bi~t
<=> phuang trlnh hoanh d9 giao ~iem f(x)
Gici su ham sob~c ba y
, , {y <small>1 </small>
y 2 - f x2 - Ax2 + B
Suy ra, cac diem (x<sub>1</sub>;y<sub>1</sub>),(x2;y2) n~m tren duemg th~ng y =Ax+ B.
+ Ham so co m9t ct,rc tri <=> ab ;;::: 0. + Ham so co ba ct,rc tri <=> ab < 0.
{a>O + Ham so co dung m9t Cl,fC tri va Cl,l'C tri la Cl,fC tieu <small>¢:> </small> b ;;::: 0 .
{a<O + Ham so co dung m9t Cl,fC tri va Cl,l'C tri la Cl,fC dgi <small>¢:> </small> b
{a>O + Ham so co hai Cl,fC tieu va m9t Cl,l'C d;;li <=> b < 0 .
{a<O + Ham so co m9t Cl,fC tieu va hai Cl,l'C d;;li <=> b > 0 .
GUtsuhamso y=ax<sup>4</sup>+bx<sup>2</sup>+c co 3 cvctri:
+ A, B n~m cimg ve 1 phia doi v6i trt.tc Oy <=> y' = 0 co hai nghi~m phan bi~t cling dau + A, B n~m cling ve 2 phia doi v6i tn,1.c Oy <=> y' = co hai nghi~m trcli dau
+A, B n~m cling ve 1 phia doi v6i trt,lC Ox <small>¢:> </small>y' = 0 co hai nghi~m phan bi~t va y <small>CD </small>.y <small>cr </small>> 0
Yco+Ycr>O
+A, B
A B
Yco·Ycr <0
(ap dvng khi khong nham duqc nghi~m va viet duqc phuang trlnh duemg th~ng di qua hai diem Cl,fC tri cua do thi ham so)
+ A, B n~m ve 2 phia doi v6i tn.tc Ox
<=> do thi c~t trvc Ox t;;1.i 3 diem phan bi~t
<=> phuang trlnh hoanh d9 giao ~iem f(x)
Gici su ham sob~c ba y
, , {y <small>1 </small>
y 2 - f x2 - Ax2 + B
Suy ra, cac diem (x<sub>1</sub>;y<sub>1</sub>),(x2;y2) n~m tren duemg th~ng y =Ax+ B.
+ Ham so co m9t ct,rc tri <=> ab ;;::: 0. + Ham so co ba ct,rc tri <=> ab < 0.
{a>O + Ham so co dung m9t Cl,fC tri va Cl,l'C tri la Cl,fC tieu <small>¢:> </small> b ;;::: 0 .
{a<O + Ham so co dung m9t Cl,fC tri va Cl,l'C tri la Cl,fC dgi <small>¢:> </small> b
{a>O + Ham so co hai Cl,fC tieu va m9t Cl,l'C d;;li <=> b < 0 .
{a<O + Ham so co m9t Cl,fC tieu va hai Cl,l'C d;;li <=> b > 0 .
GUtsuhamso y=ax<sup>4</sup>+bx<sup>2</sup>+c co 3 cvctri:
- - a -b<small>3 </small>
DiH a=BAC :::>coe-=-.
Tam giac ABC vuong can <small>t~i </small>A
<small>~~·~~~~···~···"~···~~~·~~····~·~ </small>
Tam giac ABC c6 3 g6c nhQn
Tr1;1c hoanh chia tam giac ABC thanh hai haiphan c6 <small>di~n </small>tich nhau
Tam giac ABC c6 diem c\].'c tri each deu Ox Do thi ham
+ bx<small>2 </small>
+ c c~t tr1;1c Ox tCJ.i 4 diem phan <small>th~mh </small>
Hlnh ph~ng gioi h9.n boi do thi
phan tren va phan duoi ox b~ng nhau
b<small>2 </small>=8ac b2 <small>= </small>100 ac
Tam giac ABC vuong can <small>t~i </small>A
<small>~~·~~~~···~···"~···~~~·~~····~·~ </small>
Tam giac ABC c6 3 g6c nhQn
Tr1;1c hoanh chia tam giac ABC thanh hai haiphan c6 <small>di~n </small>tich nhau
Tam giac ABC c6 diem c\].'c tri each deu Ox Do thi ham
+ bx<small>2 </small>
+ c c~t tr1;1c Ox tCJ.i 4 diem phan <small>th~mh </small>
Hlnh ph~ng gioi h9.n boi do thi
phan tren va phan duoi ox b~ng nhau
b<small>2 </small>=8ac b2 <small>= </small>100 ac
<small>9 </small>
<i><small>X </small></i>
</div><span class="text_page_counter">Trang 16</span><div class="page_container" data-page="16">Tlm m de do thi ham so y <small>= </small>x <small>-</small> mx + co ba diem cvc tri Iap thanh m()t tam ghic
Ba diem Cl,l'C tri Iap thanh m()t tam giac vuong thl b<small>3 </small>
= -8a <=> -m<sup>3 </sup>= -8 <=> m = 2. So gia tri nguyen cua tham so mde do thi cua ham so y = x<small>4 </small>
<small>-</small> 2 ( m +
Di'eu ki~n de' do thi ham trung phuong y = ax<small>4 </small>
+ bx<small>2 </small>+ c co ba diem cvc tri la ab < 0 <=> m > -1 Khi do ba diem Cl,l'C tri Iap thanh tam giac vuong can khi b<small>3 </small>
+ 8a = 0 <=> -8( m + 1
Di'eu ki~n de' do thi ham trung phuong y = ax<small>4 </small>+ bx<small>2 </small>
+ c co ba die'm cvc tri la ab < 0 <=> m > 0 . Khi do ba die'm cvc tri t9-o thanh tam giac co di~n tich b~ng 3 khi
32.3<sup>3</sup>.3<sup>2 </sup>+(-2m
So gia tri thvc cua tham so m de do thi ham so y = x<small>4 -</small> 2mx<small>2 </small>
+ m -1 co ba diem Cl,l'C tri t9-o thanh m()t tam giac co ban kfnh du&ng tron ngo9-i tiep b~ng 1 la ?
Di'eu ki~n de ham so co 3 diem cvc tri la: 1( -2m)<
Ap dvng cong thuc ba diem Cl,l'C tri ham hung phuong tgo thanh m()t tam giac co ban kinh du&ng , . '"" , b' k'nh R b3 -8a 1 (-2mf -8
tron ngo9.1 hep co an 1 = <small>-~-~ </small>- <small><::> </small> = ( )
<small>-</small> 2 ( m +
Di'eu ki~n de' do thi ham trung phuong y = ax<small>4 </small>
+ bx<small>2 </small>+ c co ba diem cvc tri la ab < 0 <=> m > -1 Khi do ba diem Cl,l'C tri Iap thanh tam giac vuong can khi b<small>3 </small>
+ 8a = 0 <=> -8( m + 1
Di'eu ki~n de' do thi ham trung phuong y = ax<small>4 </small>+ bx<small>2 </small>
+ c co ba die'm cvc tri la ab < 0 <=> m > 0 . Khi do ba die'm cvc tri t9-o thanh tam giac co di~n tich b~ng 3 khi
32.3<sup>3</sup>.3<sup>2 </sup>+(-2m
So gia tri thvc cua tham so m de do thi ham so y = x<small>4 -</small> 2mx<small>2 </small>
+ m -1 co ba diem Cl,l'C tri t9-o thanh m()t tam giac co ban kfnh du&ng tron ngo9-i tiep b~ng 1 la ?
Di'eu ki~n de ham so co 3 diem cvc tri la: 1( -2m)<
Ap dvng cong thuc ba diem Cl,l'C tri ham hung phuong tgo thanh m()t tam giac co ban kinh du&ng , . '"" , b' k'nh R b3 -8a 1 (-2mf -8
tron ngo9.1 hep co an 1 = <small>-~-~ </small>- <small><::> </small> = ( )
</div><span class="text_page_counter">Trang 17</span><div class="page_container" data-page="17">Co bao nhieu gia tri nguyen cua tham
y = lmx<small>3 -</small> 3mx<small>2 </small>
+(3m- 2)x + 2- ml co 5 diem cvc tri?
Xet ham
+ (3m-2) x +2-m .
Taco: mx<small>3</small>-3mx<small>2</small>
+(3m-2)x+2-m=O <=>[x=<sub>2</sub>l ( )' mx -2mx+m-2 =0 1
Yeu cau bai toan <small>¢::> </small>phuang trlnh f (X) = 0 co ba nghi~m phan bi~t <small>¢::> </small>phuang trlnh ( 1) co hai
{m<small>2 </small>
-m(m-2) > 0 <small>nghi~m </small>phan <small>bi~t </small>khac 1 <=> .
m-2m+m-2 <i><small>=F </small></i>0 Vl m nguyen va mE [ -10;10] nen mE {1;2; ... ;10}.
<small>-</small> 9x- 5 + m <small>2</small>
<small>1 </small>co 5 diem
Ve do thi ham
<small>-</small> 9x-
Ham
cling luon co 2 diem cvc tri. Do do yeu cau bai to an <small>¢::> </small>
32 <small>2 </small> 0 {m<sup>2 </sup>< 32
<=> - < -m < <=> <=> . m:t=O <i><small>m=F-0 </small></i>
V6i mE
+(3m- 2)x + 2- ml co 5 diem cvc tri?
Xet ham
+ (3m-2) x +2-m .
Taco: mx<small>3</small>-3mx<small>2</small>
+(3m-2)x+2-m=O <=>[x=<sub>2</sub>l ( )' mx -2mx+m-2 =0 1
Yeu cau bai toan <small>¢::> </small>phuang trlnh f (X) = 0 co ba nghi~m phan bi~t <small>¢::> </small>phuang trlnh ( 1) co hai
{m<small>2 </small>
-m(m-2) > 0 <small>nghi~m </small>phan <small>bi~t </small>khac 1 <=> .
m-2m+m-2 <i><small>=F </small></i>0 Vl m nguyen va mE [ -10;10] nen mE {1;2; ... ;10}.
<small>-</small> 9x- 5 + m <small>2</small>
<small>1 </small>co 5 diem
Ve do thi ham
<small>-</small> 9x-
Ham
cling luon co 2 diem cvc tri. Do do yeu cau bai to an <small>¢::> </small>
32 <small>2 </small> 0 {m<sup>2 </sup>< 32
<=> - < -m < <=> <=> . m:t=O <i><small>m=F-0 </small></i>
V6i mE
(2m-1) -3(2-m)>O .1.>0
P>O <sub>2-m </sub><sup>3 </sup>- - > 0
g' ( x) = 0 ~ , ( <sub>3 </sub> <sub>2</sub><small>) _ <:::> </small> x + 3x = a, (a < 0) £ x + 3x - o <sub>x + x = , < < </sub><small>3 </small> 3 <small>2 </small> b (o b 4)
x<small>3 </small>+ 3x<small>2 </small>
Dl,l'a vao bang bien thlen tren ta thay: Phuong trlnh x<small>3 </small>
+ 3x<small>2 </small>
= a, (a > 0) c6 m{)t nghi~m don x = x<sub>1 </sub>< -2 Phuong trlnh x<small>3 </small>
(2m-1) -3(2-m)>O .1.>0
P>O <sub>2-m </sub><sup>3 </sup>- - > 0
g' ( x) = 0 ~ , ( <sub>3 </sub> <sub>2</sub><small>) _ <:::> </small> x + 3x = a, (a < 0) £ x + 3x - o <sub>x + x = , < < </sub><small>3 </small> 3 <small>2 </small> b (o b 4)
x<small>3 </small>+ 3x<small>2 </small>
Dl,l'a vao bang bien thlen tren ta thay: Phuong trlnh x<small>3 </small>
+ 3x<small>2 </small>
= a, (a > 0) c6 m{)t nghi~m don x = x<sub>1 </sub>< -2 Phuong trlnh x<small>3 </small>
</div><span class="text_page_counter">Trang 19</span><div class="page_container" data-page="19"><small>0 </small>
+ Cach 1: Dung dgo ham tgi die'm Ian <small>c~n </small>x<sub>0 </sub>
- N eu f <small>I ( </small>Xo + <i>011) </i>.£' ( xo - 0 <i><small>I </small></i>1) < 0 thl ham so dgt eve tri tgi X = Xo .
<i><small>N </small></i>
f xo <i>-011 </i>< 0
<i><small>N </small></i>
<small>+ </small><i>Buac 2: </i><small>L~p </small>bang bien thien va
<i>+ Buac 1: </i>
* Ham so da cho y = <small>f </small>(X) xac dinh va lien h,lc tren dogn
* T1m cite die'm <i>XII x21 '"I </i>xn tren khoang (a; b) <i>I </i>tgi do f<small>1 </small>
+ Cach 1: Dung dgo ham tgi die'm Ian <small>c~n </small>x<sub>0 </sub>
- N eu f <small>I ( </small>Xo + <i>011) </i>.£' ( xo - 0 <i><small>I </small></i>1) < 0 thl ham so dgt eve tri tgi X = Xo .
<i><small>N </small></i>
f xo <i>-011 </i>< 0
<i><small>N </small></i>
<small>+ </small><i>Buac 2: </i><small>L~p </small>bang bien thien va
<i>+ Buac 1: </i>
* Ham so da cho y = <small>f </small>(X) xac dinh va lien h,lc tren dogn
* T1m cite die'm <i>XII x21 '"I </i>xn tren khoang (a; b) <i>I </i>tgi do f<small>1 </small>
</div><span class="text_page_counter">Trang 20</span><div class="page_container" data-page="20">Bcli toan : Tim min, max cua ham so y
Phuong phap: Su dvng nh?p ham so f ( x); ch<;>n Start: a; End: b; Step: b
Chu
Gia tri nh6 nhat cua ham so f
Cho ham so y = f(x) xac dinh tren m<)t khoang vo hq.n (la khoang dq.ng (a; +oo) <i><small>I ( </small></i>-oo; b) ho~c ( -oo; +oo) ). Duang th~ng y =Yo 1a duang ti~m c~n ngang (hay ti~m c?n ngang) cua do thi ham so y = f( neu it nhat m<)t trong cac di'eu ki~n sau duqc tho a man:
lim f ( x) =Yo, lim f ( x) =Yo
TA? <sub>ong so uong </sub><sup>Nd ' </sup> <sub>ti~m </sub><sup>•A </sup> <sub>cq.n ung va </sub><sup>A d' </sup> <sup>' oA </sup><sub>h~m </sub><sub>cq.n ngang cua o t . am so </sub><sup>A </sup> <sup>? d"' hih' </sup> <i><small>N </small></i> <sub>y </sub>
Bcli toan : Tim min, max cua ham so y
Phuong phap: Su dvng nh?p ham so f ( x); ch<;>n Start: a; End: b; Step: b
Chu
Gia tri nh6 nhat cua ham so f
Cho ham so y = f(x) xac dinh tren m<)t khoang vo hq.n (la khoang dq.ng (a; +oo) <i><small>I ( </small></i>-oo; b) ho~c ( -oo; +oo) ). Duang th~ng y =Yo 1a duang ti~m c~n ngang (hay ti~m c?n ngang) cua do thi ham so y = f( neu it nhat m<)t trong cac di'eu ki~n sau duqc tho a man:
lim f ( x) =Yo, lim f ( x) =Yo
TA? <sub>ong so uong </sub><sup>Nd ' </sup> <sub>ti~m </sub><sup>•A </sup> <sub>cq.n ung va </sub><sup>A d' </sup> <sup>' oA </sup><sub>h~m </sub><sub>cq.n ngang cua o t . am so </sub><sup>A </sup> <sup>? d"' hih' </sup> <i><small>N </small></i> <sub>y </sub>
de tim <small>ti~m c~n </small>ngang
CALC v6i x=-10<sup>6 </sup>de tim <small>ti~m c~n </small>ngang
<small>V~y </small>do
CALC v6i x <small>= </small>10<sup>6 </sup><small>ti~m c~n </small>ngang
CALC v6i 1
de tim <small>ti~m c~n </small>ngang
CALC v6i x=-10<sup>6 </sup>de tim <small>ti~m c~n </small>ngang
<small>V~y </small>do
CALC v6i x <small>= </small>10<sup>6 </sup><small>ti~m c~n </small>ngang
CALC v6i 1
</div><span class="text_page_counter">Trang 22</span><div class="page_container" data-page="22">* Cach ve:
+ Gift nguyen ph'an do thi ben phai Oy cua do thi
+ B6 phan do thi ben trcii Oy cua
+ B6 phan do thi cua
(c'):y=lxl3 -31xl+1
Ve do thi
+ <i>Gift nguyen phan do thi phia tren Ox </i>~ua do thi (C): y = f
+ <i>B6 phan do thi phia du6i Ox cua (C), lay doi x-ung phan do thi bi b6 qua Ox. </i>
Ve do thi y = lx<sup>3 </sup>- 3x + 11. <i><small>y </small></i>
-3x+1( C)
+ B6 phan do thi cua (C) du6i Ox, gift nguyen (C) phia tren Ox.
+ Lay doi x{mg phan do thi bi b6 qua Ox .
v6i dfilng: y = lf(lxl)i .ta Ian luQ't bien doi <small>2 </small>do thi y = f(lxl) va y = 1£(x
Cho ham so y = f ( x) , c6 do thi (C).
<small>-</small> <i>3x </i>+ 11
Tiep tuyen cia do thi (C) tiili die'm M<sub>0 ( </sub>x<sub>0</sub> y <sub>0</sub> e (C) c6 diiing: y = y' ( x<small>0 ) ( </small>x-x<small>0 ) </small>+ y <sub>0 • </sub>
<small>Mo ( Xo; </small>y
DO !hi (C) <small>vii ( </small>
* Cach ve:
+ Gift nguyen ph'an do thi ben phai Oy cua do thi
+ B6 phan do thi ben trcii Oy cua
+ B6 phan do thi cua
(c'):y=lxl3 -31xl+1
Ve do thi
+ <i>Gift nguyen phan do thi phia tren Ox </i>~ua do thi (C): y = f
+ <i>B6 phan do thi phia du6i Ox cua (C), lay doi x-ung phan do thi bi b6 qua Ox. </i>
Ve do thi y = lx<sup>3 </sup>- 3x + 11. <i><small>y </small></i>
-3x+1( C)
+ B6 phan do thi cua (C) du6i Ox, gift nguyen (C) phia tren Ox.
+ Lay doi x{mg phan do thi bi b6 qua Ox .
v6i dfilng: y = lf(lxl)i .ta Ian luQ't bien doi <small>2 </small>do thi y = f(lxl) va y = 1£(x
Cho ham so y = f ( x) , c6 do thi (C).
<small>-</small> <i>3x </i>+ 11
Tiep tuyen cia do thi (C) tiili die'm M<sub>0 ( </sub>x<sub>0</sub>y <sub>0</sub> e (C) c6 diiing: y = y' ( x<small>0 ) ( </small>x-x<small>0 ) </small>+ y <sub>0 • </sub>
<small>Mo ( Xo; </small>y
DO !hi (C) <small>vii ( </small>
Bai to an: Viet phuang trlnh tiep tuyen t~i diem M ( x<sub>0</sub> y <sub>0 </sub><small>) </small>
Phuong trlnh tiep tuyen co d~ng d: y = kx + <small>m; </small>ta se tlm k, m nhu sau: + B1: Tlm
vay phuang trinh tiep tuyen la: y = 3x- 4 .
Tiep tuyen cua do thi
Ta co phuang trlnh: -4x<sup>3 </sup>+ 3x + 1 <small>= </small>-9x + 7 co 3 nghi~m phan bi~t ,lo~i A. Ta co phuang trlnh: -4x<small>3 </small>
+ 3x + 1 <small>= </small>-9x -7 co 2 nghi~m phan bi~t;
Cho ham so y
Phuong trlnh hoanh d9 giao diem cia (C<sub>1</sub> va (C<sub>2 ) </sub>la f(x) = g(x) (1). Khi d6: +
Bai to an: Viet phuang trlnh tiep tuyen t~i diem M ( x<sub>0</sub> y <sub>0 </sub><small>) </small>
Phuong trlnh tiep tuyen co d~ng d: y = kx + <small>m; </small>ta se tlm k, m nhu sau: + B1: Tlm
vay phuang trinh tiep tuyen la: y = 3x- 4 .
Tiep tuyen cua do thi
Ta co phuang trlnh: -4x<sup>3 </sup>+ 3x + 1 <small>= </small>-9x + 7 co 3 nghi~m phan bi~t ,lo~i A. Ta co phuang trlnh: -4x<small>3 </small>
+ 3x + 1 <small>= </small>-9x -7 co 2 nghi~m phan bi~t;
Cho ham so y
Phuong trlnh hoanh d9 giao diem cia (C<sub>1</sub> va (C<sub>2 ) </sub>la f(x) = g(x) (1). Khi d6: +
D"' hi <sub>o t . </sub>
va B, khi do do dai do9-n AB b~ng?
<i>Phuang phap 1: Bang bien thien </i>
+ L?p phuong trlnh hoanh d9 giao die'm d9-ng F ( x, m) <small>= </small>0 (phuong trlnh an <small>X </small>tham so m) + Co I?p m dua phuong trlnh ve d9-ng m
+ L?p bang bien thien cho ham so y
+ Dt.ra va gia thiet va bang bien thien ill do suy ra m.
<i>Phuang phap 2: </i>
+ Co l?p m ho~c dua ve ham h~ng la duemg thclng vuong goc voi tr1,1c Oy + Tu do thi ham so tlm C\fC d9-i, C\fC tieu cua ham so (neu co)
+ Dt.ra vao so iao diem cua hai do thi ham so ta tlm duqc gia tri cua m thea yeu cau cua bai toan Cho ham so y
ve ben du6i. Tlm so nghi~m cua phuong trlnh f (X+ 2020) = 2 .
4.
D~t t = X+ 2020 => f ( t) = 2 va voi m6i nghi~m t se cho duy nhat mot nghi~m <small>X . </small>Do do so nghi~m CUa phuong trinh f (X + 2020) = 2 cling la SO nghi~m CUa phuong trinh f ( t) = 2 va la SO giao diem CUa do thi ham SO y <small>= </small>f
Quan sat do thi ta thay co 2 giao die'm. V?y phuong trlnh da cho co 2 Cho ham so y
va B, khi do do dai do9-n AB b~ng?
<i>Phuang phap 1: Bang bien thien </i>
+ L?p phuong trlnh hoanh d9 giao die'm d9-ng F ( x, m) <small>= </small>0 (phuong trlnh an <small>X </small>tham so m) + Co I?p m dua phuong trlnh ve d9-ng m
+ L?p bang bien thien cho ham so y
+ Dt.ra va gia thiet va bang bien thien ill do suy ra m.
<i>Phuang phap 2: </i>
+ Co l?p m ho~c dua ve ham h~ng la duemg thclng vuong goc voi tr1,1c Oy + Tu do thi ham so tlm C\fC d9-i, C\fC tieu cua ham so (neu co)
+ Dt.ra vao so iao diem cua hai do thi ham so ta tlm duqc gia tri cua m thea yeu cau cua bai toan Cho ham so y
ve ben du6i. Tlm so nghi~m cua phuong trlnh f (X+ 2020) = 2 .
4.
D~t t = X+ 2020 => f ( t) = 2 va voi m6i nghi~m t se cho duy nhat mot nghi~m <small>X . </small>Do do so nghi~m CUa phuong trinh f (X + 2020) = 2 cling la SO nghi~m CUa phuong trinh f ( t) = 2 va la SO giao diem CUa do thi ham SO y <small>= </small>f
Quan sat do thi ta thay co 2 giao die'm. V?y phuong trlnh da cho co 2 Cho ham so y
Ta c6 21£ ( x
Khi d6 duemg
Phuong phap 1: Nham <small>nghi~m </small>ket hqp v6i cac phuang phap gic\i toan tam thuc <small>b~c </small>2
Phuong phap 2: Cl,l'C tri (khi bai toan khong co <small>l~p </small>duqc m va cling khong nham duqc <small>nghi~ml </small>xem them
C6 bao nhieu gia tri nguyen cua tham so m de' do thi cua ham so y =x<small>3 </small>
<i><small>::f::.O </small></i>
Cac gia tri nguyen clia m tho a man yeu cau <small>b~ti </small>toan la: 0 <i><small>I </small>11 </i>2 .
Phuong phap: Ta se dua bai toan ve cac d0ng <i>3<small>1 </small></i>chu
x 1
Ta c6 21£ ( x
Khi d6 duemg
Phuong phap 1: Nham <small>nghi~m </small>ket hqp v6i cac phuang phap gic\i toan tam thuc <small>b~c </small>2
Phuong phap 2: Cl,l'C tri (khi bai toan khong co <small>l~p </small>duqc m va cling khong nham duqc <small>nghi~ml </small>xem them
C6 bao nhieu gia tri nguyen cua tham so m de' do thi cua ham so y =x<small>3 </small>
<i><small>::f::.O </small></i>
Cac gia tri nguyen clia m tho a man yeu cau <small>b~ti </small>toan la: 0 <i><small>I </small>11 </i>2 .
Phuong phap: Ta se dua bai toan ve cac d0ng <i>3<small>1 </small></i>chu
x 1
</div><span class="text_page_counter">Trang 26</span><div class="page_container" data-page="26">Ta c6: AB = 4 <=> AB<small>2 </small>
= 16 <=> (xB
-2m-3= 0 <::>
-4x+2
<small>-</small> 2 <small>= </small>0 ~ x <small>= </small>
Phuong ha 2: <small>D~t </small>an hv ket hqp v6i cac huan hap giai toan tam thuc bac 2
So gia tri nguyen cua tham so m de duemg th~ng y = -1 c~t do thi ham so y
De' duemg th~ng y = -1 c~t do thi ham so tc~ti bon diem pharr bi~t c6 horum d9 nho han 2 thl di'eu
1 0<3m+1<4 --<m<1
<small>ki~n </small>la <small>~ </small>3 . vay khong c6 gia tri nguyen m thoa man. 3m+1:;t:1
0 m:;t: eho ham so y <small>= </small>x<small>4 </small>
<small>, ( </small>t <i><small>:?: </small></i>0).
Phuong tdnh thrum
(em) c~t Ox tiili bon diem pharr bi~t khi ( 1) c6 hai nghi~m duang phan bi~t.
=> T = ( -oo; 0) . vay c6 9 gia tri nguyen
Ta c6: AB = 4 <=> AB<small>2 </small>
= 16 <=> (xB
-2m-3= 0 <::>
-4x+2
<small>-</small> 2 <small>= </small>0 ~ x <small>= </small>
Phuong ha 2: <small>D~t </small>an hv ket hqp v6i cac huan hap giai toan tam thuc bac 2
So gia tri nguyen cua tham so m de duemg th~ng y = -1 c~t do thi ham so y
De' duemg th~ng y = -1 c~t do thi ham so tc~ti bon diem pharr bi~t c6 horum d9 nho han 2 thl di'eu
1 0<3m+1<4 --<m<1
<small>ki~n </small>la <small>~ </small> 3 . vay khong c6 gia tri nguyen m thoa man. 3m+1:;t:1
0 m:;t: eho ham so y <small>= </small>x<small>4 </small>
<small>, ( </small>t <i><small>:?: </small></i>0).
Phuong tdnh thrum
(em) c~t Ox tiili bon diem pharr bi~t khi ( 1) c6 hai nghi~m duang phan bi~t.
=> T = ( -oo; 0) . vay c6 9 gia tri nguyen
x=1 <small>V~y </small>co hai giao diem.
G . M N 1' <sub>91 </sub> <sub>, </sub> <sub>a cac g1ao 1em cua </sub>' . d'"' ' h . d"' h' h' <sub>m o t </sub><sub>l </sub> <sub>am so y = x-</sub>"' 2 <sub>va y = </sub>' <sup>7</sup>x-<sup>14 </sup>2 GQi I 1a trung die'm cua do9-n th~ng MN. Tim hoanh d<) die'm I.
7
2 x+
Xet phuang trinh hoanh d<) giao diem: x- 2 = <sup>7</sup>x 214
-. Su dt,mg SHIFT +SOLVE hoi;ic dua ve d9-ng x+
I 1, <sup>d' "' </sup> <sup>' d </sup> <sup>h_, </sup> <sup>MN " </sup> <sup>' </sup> <sup>2 </sup>+ <sup>5 7 </sup>Do a trung 1em cua 09-n t ang nen ta co x<sub>1 </sub>= -
2<sup>- =- . </sup>
' h 'nhh 'nhd" · d'"' 2 1 <sup>2</sup>x-<sup>1 </sup> h' d' ' A ,. 2 h"' " Xet p uang tn oa Q g1ao 1em mx+m+ = - - ; t u ap <small>an </small> va1 m= t ay vo
2x+1 ngJ:t:~~~ A. Thu d~p ~n C v6i m = -1 tl-l~Y ~<) 2 ng~~_I!l v~y
Bu6c tinh <small>Nh~p </small>
2x-1 2mx+m+1=--
2x+1
Tim <small>nghi~m </small>v6i m = 2
Tim <small>nghi~m </small>v6i m = -1
Xet phuang tdnh hoanh d<) giao diem: x - 3x + 2 = 0 <small>--=~'"---7'> </small> . x=1 <small>V~y </small>co hai giao diem.
G . M N 1' <sub>91 </sub> <sub>, </sub> <sub>a cac g1ao 1em cua </sub>' . d'"' ' h . d"' h' h' <sub>m o t </sub><sub>l </sub> <sub>am so y = x-</sub>"' 2 <sub>va y = </sub>' <sup>7</sup>x-<sup>14 </sup>2 GQi I 1a trung die'm cua do9-n th~ng MN. Tim hoanh d<) die'm I.
7
2 x+
Xet phuang trinh hoanh d<) giao diem: x- 2 = <sup>7</sup>x 214
-. Su dt,mg SHIFT +SOLVE hoi;ic dua ve d9-ng x+
I 1, <sup>d' "' </sup> <sup>' d </sup> <sup>h_, </sup> <sup>MN " </sup> <sup>' </sup> <sup>2 </sup>+ <sup>5 7 </sup>Do a trung 1em cua 09-n t ang nen ta co x<sub>1 </sub>= -
2<sup>- =- . </sup>
' h 'nhh 'nhd" · d'"' 2 1 <sup>2</sup>x-<sup>1 </sup> h' d' ' A ,. 2 h"' " Xet p uang tn oa Q g1ao 1em mx+m+ = - - ; t u ap <small>an </small> va1 m= t ay vo
2x+1 ngJ:t:~~~ A. Thu d~p ~n C v6i m = -1 tl-l~Y ~<) 2 ng~~_I!l v~y
Bu6c tinh <small>Nh~p </small>
2x-1 2mx+m+1=--2x+1
Tim <small>nghi~m </small>v6i m = 2
Tim <small>nghi~m </small>v6i m = -1
</div><span class="text_page_counter">Trang 28</span><div class="page_container" data-page="28"><small>T~p </small>Xck dinh cua ham
·:· T~p xac dinh cua ham
T~p xac d:inh: ( 0;
y' = a.xa-<small>1 </small>
> 0 Vx > 0. lim X a = <i>01 </i> lim X a = +oo,
<small>x~o+ x~+«> </small>
y=xa ,a< 0. T~p xac dinh: ( 0;
y'=a.xa-<small>1 </small>
<0 Vx>O limxa =+oo, limxa =0.
<small>x~o+ x~+«> </small>
Khong co <small>ti~m c~n. </small>Bang bien thien.
Ox la <small>ti~m c~n </small>ngang; Oy la <small>ti~m c~n </small>dling. Bang bien thien.
<small>X 0 </small>
<small>+ </small>
Do thj ham
Khaos;:ithan1somi:i y=ax (a>O,a;t=l)
<small>T~p </small>Xck dinh cua ham
·:· T~p xac dinh cua ham
T~p xac d:inh: ( 0;
y' = a.xa-<small>1 </small>
> 0 Vx > 0. lim X a = <i>01 </i> lim X a = +oo,
<small>x~o+ x~+«> </small>
y=xa ,a< 0. T~p xac dinh: ( 0;
y'=a.xa-<small>1 </small>
<0 Vx>O limxa =+oo, limxa =0.
<small>x~o+ x~+«> </small>
Khong co <small>ti~m c~n. </small>Bang bien thien.
Ox la <small>ti~m c~n </small>ngang; Oy la <small>ti~m c~n </small>dling. Bang bien thien.
<small>X 0 </small>
<small>+ </small>
Do thj ham
Khaos;:ithan1somi:i y=ax (a>O,a;t=l)
</div><span class="text_page_counter">Trang 29</span><div class="page_container" data-page="29"><i><small>\j </small></i>
<small>il < J I </small>
Cho hai
a= loga b <=> aa = b, (a,b > O,a <i><small>-=1= </small></i>1)
loga 1 =
loga ba = a.loga
logaP ba = p.loga b
log. b -log. c
loga b =log b loga <small>C c </small>
logaP b =
1 log b = - -
<i><small>\j </small></i>
<small>il < J I </small>
Cho hai
a= loga b <=> aa = b, (a,b > O,a <i><small>-=1= </small></i>1)
loga 1 =
loga ba = a.loga
logaP ba = p.loga b
log. b -log. c
loga b =log b loga <small>C c </small>
logaP b =
1 log b = - -
<small>a> I </small>
<i><small>X </small></i>
<small>O<a<l </small>
</div><span class="text_page_counter">Trang 30</span><div class="page_container" data-page="30"><small>V~y </small>a=11;b=8=>a<sup>2</sup>-b<sup>3 </sup>=-391.
<b>Phuong phap </b>3. <small>D~t </small>an ph1,1 + Lo9-i 1: m.a<small>2</small>
f(x) +n.af(x) +p = 0 me +nt+p=O + Lo9-i 2: f ( ag(x)) <small>t=ag(xl </small> f ( t) = 0
+ Lo9-i 3: m.a2f(x) + n.( a.b y(x) + p.b2f(x) = 0~ Chia ca 2
Tong tat ca cac <small>nghi~m </small>cua phuang trlnh 9x - 4.3x + 3 = 0 la
Ta c6: gx - 4.3x + 3 = 0 <small>¢:> [</small>3
x = <sup>1 </sup><small>¢:> </small>
Tinh tong T tat ca cac <small>nghi~m </small>clia phuang trlnh 4.9x -13.6x +9.4x
<small>V~y a=11;b=8=>a</small>2<sub>-b</sub>3 =-391.
<b>Phuong phap </b>3. <small>D~t </small>an ph1,1 + Lo9-i 1: m.a<small>2</small>
f(x) +n.af(x) +p = 0 me +nt+p=O + Lo9-i 2: f ( ag(x)) <small>t=ag(xl </small> f ( t) = 0
+ Lo9-i 3: m.a2f(x) + n.( a.b y(x) + p.b2f(x) = 0~ Chia ca 2
Tong tat ca cac <small>nghi~m </small>cua phuang trlnh 9x - 4.3x + 3 = 0 la
Ta c6: gx - 4.3x + 3 = 0 <small>¢:> [</small>3
x = <sup>1 </sup><small>¢:> </small>
Tinh tong T tat ca cac <small>nghi~m </small>clia phuang trlnh 4.9x -13.6x +9.4x
6x +3 2x 6x +(3+m)2x +m <small>= </small>0 <small>Â::> ã </small> = -m.
2x +1 D<sub>;;tt </sub><small>v </small>
2<sup>x + </sup>1 <sup>VOl X E </sup> <sup>; </sup> <sup>. </sup>
<small>, 1 </small>
<small>( </small>6x ln6+3.2x ln2 )( 2x + 1 )-( 6x +3.2x )( 2x ln2) Taco f (
<small>2 </small>
6x2x (ln6 -ln2 )+ 6x ln6 + 3.2x ln2 [
<small>-~---''---</small>2 <small>- - - -</small>> 0 , <small>V'x E </small> 0; 1 ( 2x + 1)
Suy ra f (
6x +3 2x 6x +(3+m)2x +m <small>= </small>0 <small>Â::> ã </small> = -m.
2x +1 D<sub>;;tt </sub><small>v </small>
2<sup>x + </sup>1 <sup>VOl X E </sup> <sup>; </sup> <sup>. </sup>
<small>, 1 </small>
<small>( </small>6x ln6+3.2x ln2 )( 2x + 1 )-( 6x +3.2x )( 2x ln2) Taco f (
<small>2 </small>
6x2x (ln6 -ln2 )+ 6x ln6 + 3.2x ln2 [
<small>-~---''---</small>2 <small>- - - -</small>> 0 , <small>V'x E </small> 0; 1 ( 2x + 1)
Suy ra f (
<b>Phuong phap 1. </b>Dua
{0 <a <i><small>=1= </small></i>1 loga f(x) = loga g(x) <=> f{x) = g(x) > <sub>0 </sub>
DK: 0<x<5; log<sub>6 </sub> x(s-x) =1 <=> x -5x+6=0 <=> x=3
Di'eu <small>ki~n: </small>x > 3
log<sub>2 </sub><small>( </small>x- 3) + log<sub>2 ( </sub>x -1) = 3 <=> log<sub>2 ( </sub>x- 3 )( x -1) = 3
dfeu <small>ki~n </small>suy ra phuong trlnh c6 <small>nghi~m </small>x = 5 .
Tong tat ca cac nghi~m cua phuong trlnh log<sub>2 </sub><small>( </small>3.2x
<b>Phuong phap 2. </b>D~t an phv: f (log <small>a </small>g ( x)) = 0 t=log. g(x) ) f (
DK: 0<x<5; log<sub>6 </sub> x(s-x) =1 <=> x -5x+6=0 <=> x=3
Di'eu <small>ki~n: </small>x > 3
log<sub>2 </sub><small>( </small>x- 3) + log<sub>2 ( </sub>x -1) = 3 <=> log<sub>2 ( </sub>x- 3 )( x -1) = 3
dfeu <small>ki~n </small>suy ra phuong trlnh c6 <small>nghi~m </small>x = 5 .
Tong tat ca cac nghi~m cua phuong trlnh log<sub>2 </sub><small>( </small>3.2x
<b>Phuong phap 2. </b>D~t an phv: f (log <small>a </small>g ( x)) = 0 t=log. g(x) ) f (
Nh?n thay x = lla 1 <small>nghi~m. </small>Nen phuang trlnh 2x = 2 -log3 x co <small>nghi~m </small>duy nhat la x 1.
<i>Bat phuong trlnh mil ca b2m co d9ng ax> </i>b <small>(ho~c ax~ </small><i>b<small>1ax </small></i>< <i>b<small>1ax::;, </small></i>b) voi a> <i>0<small>1a </small></i><small>:;t: 1. </small>Ta xet bat phuang trlnh co d<;lng a <small>X </small>> b (
+ Neu b::;, 0 <i><small>I </small></i>t?p <small>nghi~m </small>cua bat phuang trlnh la 1R .
a> 1 x >lo b b > 0 thi ( 1) ¢:::>ax >a <small>log, b </small>¢:::> ga
{0 <a< 1 X< loga b <small>T~p nghi~m </small>cua bat phuang trlnh
Nh?n thay x = lla 1 <small>nghi~m. </small>Nen phuang trlnh 2x = 2 -log3 x co <small>nghi~m </small>duy nhat la x 1.
<i>Bat phuong trlnh mil ca b2m co d9ng ax> </i>b <small>(ho~c ax~ </small><i>b<small>1ax </small></i>< <i>b<small>1ax::;, </small></i>b) voi a> <i>0<small>1a </small></i><small>:;t: 1. </small>Ta xet bat phuang trlnh co d<;lng a <small>X </small>> b (
+ Neu b::;, 0 <i><small>I </small></i>t?p <small>nghi~m </small>cua bat phuang trlnh la 1R .
a> 1 x >lo b b > 0 thi ( 1) ¢:::>ax >a <small>log, b </small>¢:::> ga
{0 <a< 1 X< loga b <small>T~p nghi~m </small>cua bat phuang trlnh
</div><span class="text_page_counter">Trang 34</span><div class="page_container" data-page="34"><small>T~p nghi~m </small>cua bat phuong trinh 2x +4.5x -4 <lOX la
2x +4.5x -4 <lOx <small>¢:> </small>2x -lOx +4.5x -4 < 0 ¢:> 2x (1-5x )-4(1-5x) < 0
¢>(1-5x)(2x -4)<0
1- 5x < 0 {5x > 1 2x -4 > 0 2x > 4
¢> { ¢> { <=> ¢>xe(-oo;O)u(2;+oo). 1- 5x > 0 5x < 1 <small>X </small>< 0
Ket hqp v6i di'eu
Bat phuang trlnh logarit <i>ca ban c6 de:mg loga x > b </i><small>(ho~c loga x </small><i>2 </i>b,loga x < b,loga x ~b) v6i a> O,a <i><small>=t: 1. </small></i>
Bat phuang trlnh log a x > b <=> {
a >1 x>ab
{0 <a <1 0 < x <ab
So nghi~m nguyen thUQC ( -20; 20) cua bat phuang trlnh log2 ( log4
<small>T~p nghi~m </small>cua bat phuong trinh 2x +4.5x -4 <lOX la
2x +4.5x -4 <lOx <small>¢:> </small>2x -lOx +4.5x -4 < 0 ¢:> 2x (1-5x )-4(1-5x) < 0
¢>(1-5x)(2x -4)<0
1- 5x < 0 {5x > 1 2x -4 > 0 2x > 4
¢> { ¢> { <=> ¢>xe(-oo;O)u(2;+oo). 1- 5x > 0 5x < 1 <small>X </small>< 0
Ket hqp v6i di'eu
Bat phuang trlnh logarit <i>ca ban c6 de:mg loga x > b </i><small>(ho~c loga x </small><i>2 </i>b,loga x < b,loga x ~b) v6i a> O,a <i><small>=t: 1. </small></i>
Bat phuang trlnh log a x > b <=> {
a >1 x>ab
{0 <a <1 0 < x <ab
So nghi~m nguyen thUQC ( -20; 20) cua bat phuang trlnh log2 ( log4
¢::> {: > 1
log<sub>0,5 ( </sub>log<sub>2 ( </sub>2x
Bat phuang trinh 3log<sub>8 ( </sub>x + 1) -log<sub>2 ( </sub>2-x) ~ 1 c6 t~p nghi~m S
Ta c6: 3log<sub>8 ( </sub>x + 1) -log<sub>2 </sub>(2- x) ~ 1 <=> log<sub>2 ( </sub>x + 1) ~ 1 + log<sub>2 </sub>(2- x)
X + 1 ~ 2 ( 2 - X) {X ~ 1 1 ¢::> ¢::> ¢::> :::;; X < 2 .
2-x>O x<2 Khi d6 a=1, b=2.
V~y P=2a<small>2 </small>-ab+b<small>2 </small>
¢::> {: > 1
log<sub>0,5 ( </sub>log<sub>2 ( </sub>2x
Bat phuang trinh 3log<sub>8 ( </sub>x + 1) -log<sub>2 ( </sub>2-x) ~ 1 c6 t~p nghi~m S
Ta c6: 3log<sub>8 ( </sub>x + 1) -log<sub>2 </sub>(2- x) ~ 1 <=> log<sub>2 ( </sub>x + 1) ~ 1 + log<sub>2 </sub>(2- x)
X + 1 ~ 2 ( 2 - X) {X ~ 1 1 ¢::> ¢::> ¢::> :::;; X < 2 .
2-x>O x<2 Khi d6 a=1, b=2.
V~y P=2a<small>2 </small>-ab+b<small>2 </small>
</div><span class="text_page_counter">Trang 36</span><div class="page_container" data-page="36">3(log<sub>2 ( </sub>x+3)-1)
V6i f( t) =
Do d6 de bat phuong trlnh log<sub>2 </sub><small>( </small>sx -1) .log<sub>2 </sub><small>( </small>2.5x -2) 2 m c6 nghi~m v6i mQi x 2 1 thl:
Sn =A(1+rf
M6i thang gtri dung cling m9t so tien vao m9t thai gian co dinh. Cong thuc tinh: Dau m6i thang gili vao ngan hang so tien A dong v6i lai kep r% /thang thl so tien nh~n duqc ca von I3.n lai sau n thang ( n <small>E </small>N
Cong thuc tinh: Glri ngan hang so tien la A dong v6i lai suat r% /thang. M6i thang vao ngay
3(log<sub>2 ( </sub>x+3)-1)
V6i f( t) =
Do d6 de bat phuong trlnh log<sub>2 </sub><small>( </small>sx -1) .log<sub>2 </sub><small>( </small>2.5x -2) 2 m c6 nghi~m v6i mQi x 2 1 thl:
Sn =A(1+rf
M6i thang gtri dung cling m9t so tien vao m9t thai gian co dinh. Cong thuc tinh: Dau m6i thang gili vao ngan hang so tien A dong v6i lai kep r% /thang thl so tien nh~n duqc ca von I3.n lai sau n thang ( n <small>E </small>N
Cong thuc tinh: Glri ngan hang so tien la A dong v6i lai suat r% /thang. M6i thang vao ngay
</div><span class="text_page_counter">Trang 37</span><div class="page_container" data-page="37">Sn =A(1+r) -X r
Vay ngan hang so ti'en la A dong voi HH suat r%/thang. Sau dling m<)t thang
Cong thuc tinh: Cach tinh so ti'en con l<;ti sau n thang giong hoan toan cong thuc tinh g{ti ngan (1+rf -1
hang va rut tien hang thang nen taco Sn =A ( 1 + r f -X r (l+rf -1 De' sau dung n thang tra het nQ' thi sn = 0 nen A(1+rf -X = 0
r A(1+rf .r
Cong thuc tinh: Cach tinh so ti'en con l<;ti sau n thang giong hoan toan cong thuc tinh g{ti ngan (1+rf -1
hang va rut tien hang thang nen taco Sn =A ( 1 + r f -X r (l+rf -1 De' sau dung n thang tra het nQ' thi sn = 0 nen A(1+rf -X = 0
r A(1+rf .r
(1+rf -1
</div><span class="text_page_counter">Trang 38</span><div class="page_container" data-page="38">3. Jxadx=--xa+l+C(a:t=-1) a+1
4. -dx=--+C
ax = - + C
Ina xdx=sinx+C
kIna
21.
<small>I L:5. </small>ltantta:)(+bJc1x = _.!_lnlcos(ax+ b)l+c a
24.
25.
26.
<b>• Buoc 1: </b>ChQn X= <p( t) <i><small>I </small></i>trong d6 <p( t) la ham so rna ta chQn thich hqp (xem 6 bang dau hi~u) .
<b>• Buoc 2: </b>Lay vi phan hai ve: dx = <pI ( t) dt
<b>• Buoc 3: Bien doi f(x)dx theo t,dt gia su </b>
<b>• Buoc 4: </b>Khi d6 tinh:
3. Jxadx=--xa+l+C(a:t=-1) a+1
4. -dx=--+C
ax = - + C
Ina xdx=sinx+C
kIna
21.
<small>I L:5. </small>ltantta:)(+bJc1x = _.!_lnlcos(ax+ b)l+c a
24.
25.
26.
<b>• Buoc 1: ChQn X= </b><p( t) <i><small>I </small></i>trong d6 <p( t) la ham so rna ta chQn thich hqp (xem 6 bang dau hi~u) .
<b>• Buoc 2: Lay vi phan hai ve: dx = </b><pI ( t) dt
<b>• Buoc 3: Bien doi f(x)dx theo t,dt gia su </b>
<b>• Buoc 4: </b>Khi d6 tinh:
Phuong phap chung:
Di;it
Phuong phap chung:
Di;it
1J5-t
=>l=-6 -3-.t dt=-18 5t -t dt s(s-3x<sup>2</sup>
Phuong phap chung:
• Buoc 1: Ta bien doi tich phan ban dau ve dc;tng l =
· d v = f<small>2 ( </small>x) dx v =
Cach d?t u trong phuang phap tich phan tUng phan:
HQ nguyen ham cua ham so f
Phuong phap chung:
• Buoc 1: Ta bien doi tich phan ban dau ve dc;tng l =
· d v = f<small>2 ( </small>x) dx v =
Cach d?t u trong phuang phap tich phan tUng phan:
HQ nguyen ham cua ham so f