Đề kiểm tra định kỳ Giải tích 12 chương 4 trường THPT Võ Thành Trinh – An Giang

❑■➎▼ ❚❘❆ ✣➚◆❍ ❑Ý ❍➴❈ ❑Ý ■■
▼➷◆ ❚❖⑩◆ ✲ ▲❰P ✶✷
◆❣➔② ❦✐➸♠ tr❛✿. . .✴. . .✴✷✵✶✽

❙Ð ●■⑩❖ ❉Ö❈ ❱⑨ ✣⑨❖ ❚❸❖ ❆◆ ●■❆◆●

❚❘×❮◆● ❚❍P❚ ❱➹ ❚❍⑨◆❍ ❚❘■◆❍
✖✖✖✖✖✖✖✖✖✖✕

✣➲ ❝â ✷ tr❛♥❣

❚❤í✐ ❣✐❛♥ ❧➔♠ ❜➔✐✿ ✹✺ ♣❤ót

▼➣ ✤➲✿ ✶

❈❹❯ ❍➘■ ❚❘➁❈ ◆●❍■➏▼
❈➙✉ ✶✳ ❚➻♠ t➟♣ ♥❣❤✐➺♠ S ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ z + z − 2 = 0 tr➯♥ tr÷í♥❣ sè ♣❤ù❝✳
❆✳ S = {−1 − i; −1 + i}✳
❇✳ S = {1; 1 − i; 1 + i}✳
❈✳ S = {1; −1 − i; −1 + i}✳
❉✳ S = {1}✳
❈➙✉ ✷✳ ❈❤♦ sè ♣❤ù❝ z = 2 + 3i√✳ ●✐→ trà ❝õ❛ |2iz − z| ❜➡♥❣
❈✳ 113✳
❉✳ √113✳
❆✳ 15✳
❇✳ 15✳
❈➙✉ ✸✳ ❈❤♦ sè ♣❤ù❝ z t❤ä❛ ♠➣♥ |iz + 1| = 3✳ ●✐→ trà ♥❤ä ♥❤➜t ❝õ❛ |z| ❜➡♥❣
❆✳ 4✳
❇✳ 3✳
❈✳ 1✳
❉✳ 2✳

❈➙✉ ✹✳ ●å✐ z ✱ z ❧❛ ❤❛✐ ♥❣❤✐➺♠ ♣❤ù❝ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ z −2z +2 = 0✳ ❚➼♥❤ ❣✐→ trà ❝õ❛ P = z1 + z1 ✳
❆✳ P = 1✳
❇✳ P = 4✳
❈✳ P = 0✳
❉✳ P = √2✳
√
❈➙✉ ✺✳ ❈❤♦
w
=
(5
−
2i)(−3
+
2i)
✳
●✐→
trà
❝õ❛
2|w|
−
5
377 ❜➡♥❣
❆✳ −10√377✳
❇✳ 10√377✳
❈✳ 7√377✳
❉✳ −3√377✳
❈➙✉ √✻✳ ❈❤♦ sè ♣❤ù❝ z t❤ä❛ ♠➣♥ 2z − (3 + 4i) = 5 − 2i✳ √▼æ✲✤✉♥ ❝õ❛ z ❜➡♥❣ ❜❛♦ ♥❤✐➯✉❄
❇✳ 5✳
❈✳ 17✳
❉✳ √29✳

❆✳ 15✳
❈➙✉ ✼✳ ❈❤♦
sè ♣❤ù❝ z t❤ä❛ ♠➣♥ 3z − 2(z − 1) + 8 − 5i = 0✳ ❚➼♥❤ ♠æ✲✤✉♥ ❝õ❛ z ✳
❆✳ |z| = √11✳
❇✳ |z| = 121✳
❈✳ |z| = 11✳
❉✳ |z| = √101✳
❈➙✉ ✽✳ ❈❤♦ sè ♣❤ù❝ z t❤ä❛ ♠➣♥ z −√(1 − 3i)(−2 + i) = 2i✳ ❚➼♥❤
|z|✳
❈✳ |z| = √82✳
❉✳ |z| = 4√5✳
❆✳ |z| = 2✳
❇✳ |z| = 5 2✳
❈➙✉ ✾✳ ●å✐ z ✱ z ✱ z ✱ z ❧➔ ❝→❝ ♥❣❤✐➺♠ ♣❤ù❝ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ z − 4z − 27 = 0✳ ●✐→ trà ❝õ❛ ❜✐➸✉ t❤ù❝
T = |z | + |z |√+ |z | + |z | ❜➡♥❣ ❜❛♦ ♥❤✐➯✉❄
❆✳ T = 3 + 3✳
❇✳ T = 6 + 2√3✳
❈✳ T = 12✳
❉✳ T = 6 + √6✳
❈➙✉ ✶✵✳ ❙è ♣❤ù❝ ❧✐➯♥ ❤ñ♣ ❝õ❛ sè ♣❤ù❝ z = − 21 + 32 i ❧➔
❆✳ z = 32 − 21 i✳
❇✳ z = − 12 − 32 i✳
❈✳ z = 12 − 32 i✳
❉✳ z = 12 + 32 i✳
3

1

2

2

2

1

1

1

2

2

3

3

4

4

2

2

4

❈➙✉ ✶✶✳ ❈❤♦ ❤❛✐ sè ♣❤ù❝ z

1

= 9 − i ✈➔ z2 = −3 + 2i✳ ●✐→✐ trà ❝õ❛

√
2 154
✳
13

z1
❜➡♥❣ ❜❛♦ ♥❤✐➯✉❄
z2
82
✳
13

616
❆✳
❇✳ 169
✳
❈✳ 82
✳
❉✳
13
❈➙✉ ✶✷✳
❈❤♦ ❝→❝ sè ♣❤ù❝ z = 2 − 5i ✈➔ z = −2 − 3i✳ ❍➣② t➼♥❤ |z − z |✳
❇✳ 20✳
❈✳ 12✳
❉✳ 2√3✳
❆✳ 2√5✳

❈➙✉ ✶✸✳ ❈❤♦ sè ♣❤ù❝ w = 3 + 4i✳ ●✐→ trà ❝õ❛ S = 2|w| − 1 ❜➡♥❣ ❜❛♦ ♥❤✐➯✉❄
❆✳ S = 10✳
❇✳ S = 9✳
❈✳ S = 11✳
❉✳ S = 5✳
❈➙✉ ✶✹✳ ❈❤♦ sè ♣❤ù❝ z t❤ä❛ ♠➣♥ z = (3 − i)(4 − 2i)✳ ❍➣② t➼♥❤ ♠æ✲✤✉♥ ❝õ❛ sè ♣❤ù❝ w = iz − 11 −+ ii ✳
❆✳ |w| = 19✳
❇✳ |w| = √181✳
❈✳ |w| = 181✳
❉✳ |w| = √19✳
❈➙✉ ✶✺✳ ❇✐➳t r➡♥❣ sè ♣❤ù❝ ❧✐➯♥ ❤ñ♣ ❝õ❛ z ❧➔ z = (2 + 3i) + (4 − 8i)✳ ❚➻♠ sè ♣❤ù❝ z✳
❆✳ z = −6 − 5i✳
❇✳ z = 6 + 5i✳
❈✳ z = −6 + 5i✳
❉✳ z = 6 − 5i✳
❈➙✉ ✶✻✳ ❚r➯♥ ♠➦t ♣❤➥♥❣ tå❛ ✤ë Oxy✱ ✤✐➸♠ ❜✐➸✉ ❞✐➵♥ ❝õ❛ sè ♣❤ù❝ z = −5 − 6i ❧➔ ✤✐➸♠ ♥➔♦ s❛✉ ✤➙②❄
❆✳ P (5; −6)✳
❇✳ Q(5; 6)✳
❈✳ M (−5; 6)✳
❉✳ N (−5; −6)✳
1

2

1

2

❚r❛♥❣ ✶✴✷ ▼➣ ✤➲ ✶

❈➙✉ ✶✼✳ ❚➻♠
sè ♣❤ù❝ z t❤ä❛ ♠➣♥ |z| = 5 ✈➔ ♣❤➛♥ t❤ü❝ ❝õ❛ z ❣➜♣
❧➛♥ ♣❤➛♥ ↔♦ ❝õ❛
√ ❤❛✐ √
√ ♥â✳ √
❆✳ z = 2√√5 + i√√5 ❤♦➦❝ z = −2√√5 − √i√5✳
❇✳ z = −2
5
+
i
5
❤♦➦❝
z
=
2
√
√
√ 5 − i√ 5✳
❉✳ z = 5 + 2 5i ❤♦➦❝ z = − 5 − 2 5i✳
❈✳ z = − 5 + 2 5i ❤♦➦❝ z = 5 − 2 5i✳
❈➙✉ ✶✽✳ ❚r♦♥❣ ♠➦t ♣❤➥♥❣ tå❛ ✤ë Oxy✱ ❝❤♦ ✤✐➸♠ M ❧➔ ✤✐➸♠ ❜✐➸✉ ❞✐➵♥ ❝❤♦ sè ♣❤ù❝ z = a + bi✳ ❚➼♥❤
S = a + b✳

y
1

O

M

3

x

❆✳ S = 4✳
❇✳ S = 1✳
❈✳ S = 2✳
❉✳ S = 3✳
❈➙✉ ✶✾✳ ❈❤♦ sè ♣❤ù❝ z = 2 − 3i✳ ▼➺♥❤ ✤➲ ♥➔♦ s❛✉ ✤➙② ✤ó♥❣❄
❆✳ P❤➛♥ t❤ü❝ ❝õ❛ z ❜➡♥❣ 2 ✈➔ ♣❤➛♥ ↔♦ ❝õ❛ z ❜➡♥❣ 3i✳
❇✳ P❤➛♥ t❤ü❝ ❝õ❛ z ❜➡♥❣ 2 ✈➔ ♣❤➛♥ ↔♦ ❝õ❛ z ❜➡♥❣ 3✳
❈✳ P❤➛♥ t❤ü❝ ❝õ❛ z ❜➡♥❣ 2 ✈➔ ♣❤➛♥ ↔♦ ❝õ❛ z ❜➡♥❣ −3i✳
❉✳ P❤➛♥ t❤ü❝ ❝õ❛ z ❜➡♥❣ 2 ✈➔ ♣❤➛♥ ↔♦ ❝õ❛ z ❜➡♥❣ −3✳
❈➙✉ ✷✵✳ ❈❤♦ sè ♣❤ù❝ w t❤ä❛ ♠➣♥ (3 − 2i)w = 4 + 2i✳ ❚➻♠ sè ♣❤ù❝ ❧✐➯♥ ❤ñ♣ ❝õ❛ w✳
14
8
❆✳ w = 4 − 2i✳
i✳
❉✳
− i✳
❇✳ w = 3 + 2i✳
❈✳ w = 138 + 14
w=
13
13 13
✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ❍➌❚✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲

❚r❛♥❣ ✷✴✷ ▼➣ ✤➲ ✶

❇❷◆● ✣⑩P ⑩◆ ❈⑩❈ ▼❶ ✣➋
▼➣ ✤➲ t❤✐ ✶
✶✳

❈

✷✳

❉

✸✳

❉

✹✳ ❆

✺✳

❉

✻✳

❈

✼✳

❉

✽✳

❈

✾✳

❇

✶✶✳

✶✵✳
❉

✶✷✳ ❆

✶✸✳

❇

✶✹✳

✶✺✳

❇

✶✻✳

✶✼✳ ❆
✶✾✳

❇
❇

❉

✶✽✳ ❆
✷✵✳

❉

✶

❉


✣⑩P ❈❍■ ❚■➌❚ ▼❶ ✣➋ ✶
❈➙✉ ✶✳
z 3 + z 2 − 2 = 0 ⇔ (z − 1)(z 2 + 2z + 2) = 0 ⇔

z=1
z−1=0
⇔
2
z = −1 ± i.
z + 2z + 2 = 0

❱➟② t➟♣ ♥❣❤✐➺♠ ❝õ❛ ♣❤÷ì♥❣ tr➻♥❤ ❧➔ S = {1; −1 − i; −1 + i}✳
❈❤å♥ ✤→♣ →♥ ❈

❈➙✉ ✷✳ |2iz − z| = |2i(2 + 3i) − (2 − 3i)| = |−8 + 7i| =

(−8)2 + 72 =

√
113✳

❈❤å♥ ✤→♣ →♥ ❉

❈➙✉ ✸✳

✣➦t z = x + yi ✈î✐ x, y ∈ R✳
●å✐ M (x; y) ❧➔ ✤✐➸♠ ❜✐➸✉ ❞✐➵♥ ❝õ❛ sè ♣❤ù❝ z tr➯♥ ♠➦t ♣❤➥♥❣ Oxy ✳
❑❤✐ ✤â OM = |z|✳
❚❛ ❝â

y
4 Q

|iz + 1| = 3 ⇔ |xi − y + 1| = 3 ⇔ x2 + (y − 1)2 = 32 .
❙✉② r❛ M ♥➡♠ tr➯♥ ✤÷í♥❣ trá♥ t➙♠ I(0; 1) ❜→♥ ❦➼♥❤ R = 3✳ ●å✐
P (0; −2) ✈➔ Q(0; 4)✳ ❚❛ ❝â OP ≤ OM ≤ OQ✳
❙✉② r❛ OM ♥❤ä ♥❤➜t ❦❤✐ M ≡ P ✳ ❉♦ ✤â M (0; −2)✳

1

I

O

M
3

x

−2 P
❱➟② ❣✐→ trà ♥❤ä ♥❤➜t ❝õ❛ |z| ❜➡♥❣ OM =
❈❤å♥ ✤→♣ →♥ ❉

❈➙✉ ✹✳ P❤÷ì♥❣ tr➻♥❤ z

(0 − 0)2 + (−2 − 0)2 = 2✳

− 2z + 2 = 0 ❝â ∆ = (−1)2 − 1 · 2 = −1 = i2 ✳ ❱➟② ♣❤÷ì♥❣ tr➻♥❤ ❝â ❤❛✐ ♥❣❤✐➺♠
♣❤ù❝ z1 = 1 − i ✈➔ z2 = 1 + i✳
1
1
1
1
=
+
= |1| = 1✳
+
❱➟② P =
z1 z2
1−i 1+i
❈❤å♥ ✤→♣ →♥ ❆
2

❈➙✉ ✺✳ ❚❛ ❝â w = (5 − 2i)(−3 + 2i) =√−11 + 16i✳
2 + 162 =
❑❤✐ ✤â |w| =√|w| = (−11)
377√
✳

√
√
❱➟② 2|w| − 5 377 = 2 377 − 5 377 = −3 377✳
❈❤å♥ ✤→♣ →♥ ❉

❈➙✉ ✻✳

2z − (3 + 4i) = 5 − 2i ⇔ 2z = 5 − 2i + 3 + 4i ⇔ 2z = 8 + 2i ⇔ z = 4 + i.
√
❱➟② |z| = 42 + 12 = 17✳
❈❤å♥ ✤→♣ →♥ ❈
√

❈➙✉ ✼✳ ✣➦t z = x + yi ✈î✐ x, y ∈ R✳ ❑❤✐ ✤â z = x − yi✳

✷


❚❛ ❝â

3z − 2(z − 1) + 8 − 5i = 0
⇔3(x − yi) − 2(x + yi − 1) + 8 − 5i = 0
⇔3x − 3yi − 2x − 2yi + 2 + 8 − 5i = 0
⇔(x + 10) + (−5y − 5)i = 0
⇔

x + 10 = 0
− 5y − 5 = 0

⇔

x = −10
y = −1.

❈❤♦ ♥➯♥ z = −10 − i ✈➔ z =
√−10 + i✳
❱➟② |z| = (−10)2 + 12 = 101✳
❈❤å♥ ✤→♣ →♥ ❉

❈➙✉ ✽✳ ❚❛ ❝â z√− (1 − 3i)(−2
√ + i) = 2i ⇔ z = 2i + (1 − 3i)(−2 + i) ⇔ z = 1 + 9i✳
❱➟② |z| = |z| = 12 + 92 =
❈❤å♥ ✤→♣ →♥ ❈

82✳

❈➙✉ ✾✳

z = ±3
z2 = 9
√
z − 6z − 27 = 0 ⇔ 2
⇔
z = −3
z = ±i 3.
√
√
√
❱➟② T = 3 + 3 + 3 + 3 = 6 + 2 3✳
❈❤å♥ ✤→♣ →♥ ❇

4

2

❈➙✉ ✶✵✳ ❙è ♣❤ù❝ ❧✐➯♥ ❤ñ♣ ❝õ❛ sè ♣❤ù❝ z = − 12 + 32 i ❧➔ z = − 12 − 32 i✳
❈❤å♥ ✤→♣ →♥ ❇

❈➙✉ ✶✶✳

29 15
9−i
z1
=
= − − i =
z2
−3 + 2i
13 13

29
−
13

2

15
+ −
13

2

=

82
✳
13

❈❤å♥ ✤→♣ →♥ ❉

❈➙✉ ✶✷✳ |z

− z2 | = |2 − 5i − (−2 − 3i)| = |4 − 2i| =
❈❤å♥ ✤→♣ →♥ ❆
√
❚❛ ❝â |w| = 32 + 42 = 5✳
❱➟② S = 2 · 5 − 1 = 9✳
❈❤å♥ ✤→♣ →♥ ❇
1

√
42 + (−2)2 = 2 5✳

❈➙✉ ✶✸✳

❈➙✉ ✶✹✳ ❚❛ ❝â z = (3 − i)(4 − 2i) = 10 − 10i✳

1+i
❑❤✐ ✤â w = i(10 − 10i) −
= 10 + 10i − i = 10 + 9i✳
√1−i
√

❱➟② |w| = 102 + 92 = 181✳
❈❤å♥ ✤→♣ →♥ ❇

❈➙✉ ✶✺✳ ❚❛ ❝â z = (2 + 3i) + (4 − 8i) = (2 + 4) + (3 − 8)i = 6 − 5i✳
❱➟② z = z = 6 + 5i✳
❈❤å♥ ✤→♣ →♥ ❇

❈➙✉ ✶✻✳ ❚r➯♥ ♠➦t ♣❤➥♥❣ tå❛ ✤ë Oxy✱ ✤✐➸♠ ❜✐➸✉ ❞✐➵♥ ❝õ❛ sè ♣❤ù❝ z = −5 − 6i ❧➔ ✤✐➸♠ N (−5; −6)✳
❈❤å♥ ✤→♣ →♥ ❉

✸


❈➙✉ ✶✼✳ ✣➦t z = x + yi ✈î✐ x, y ∈ R✳
❚❤❡♦ ✤➲ ❜➔✐ t❛ ❝â


x = 2y
⇔
x2 + y 2 = 25



x = 2y

⇔

y2 = 5


√
x=2 5
√
y= 5
√
x = −2 5
√
y = − 5.

√
√
√
√
❱➟② z = 2 5 + i 5 ❤♦➦❝ z = −2 5 − i 5✳
❈❤å♥ ✤→♣ →♥ ❆

❈➙✉ ✶✽✳ ❚❛ ❝â M (3; 1) ♥➯♥ z = 3 + i✳ ❈❤♦ ♥➯♥ S = 3 + 1 = 4✳
❈❤å♥ ✤→♣ →♥ ❆

❈➙✉ ✶✾✳ P❤➛♥ t❤ü❝ ❝õ❛ z ❜➡♥❣ 2 ✈➔ ♣❤➛♥ ↔♦ ❝õ❛ z ❜➡♥❣ −3✳
❈❤å♥ ✤→♣ →♥ ❉

❈➙✉ ✷✵✳
(3 − 2i)w = 4 + 2i ⇔ w =

8
14
4 + 2i
⇔w=
+ i.

3 − 2i
13 13

8
14
− i✳
13 13
❈❤å♥ ✤→♣ →♥ ❉

❱➟② w =

❉✉②➺t ❇●❍

❉✉②➺t ❚ê tr÷ð♥❣
❚r÷ì♥❣ ❱➠♥ ❍ò♥❣

✹

❈❤ñ ▼î✐✱ ♥❣➔② ✶✻ t❤→♥❣ ✵✸ ♥➠♠ ✷✵✶✽

◆❣÷í✐ s♦↕♥

❈❛♦ ❚❤➔♥❤ ❚❤→✐