Ap dung bat dang thuc de giai PT He PT

<span class='text_page_counter'>(1)</span>B, ¸p dông c¸c biÓu thøc d¬ng gi¶i ph¬ng tr×nh vµ hÖ ph¬ng tr×nh: Bµi 1: Gi¶i ph¬ng tr×nh: 3 x 2  6 x  12  5 x 2  10 x  9 3  4 x  2 x 2 (*). Gi¶i: Ta cã: 3x + 6x + 12 = 3x + 6x + 3 + 9 = 3(x +1)2 + 9  9 víi mäi x. 5x2+ 10x + 9 = 5x2+ 10x + 5 + 4 = 5(x + 1)2+ 4  4 víi mäi x. 2. 2. 3 x 2  6 x  12  5 x 2  10 x  9  4  9 5 (1). . Mµ 3 - 4x - 2x2 = 5 - 4x- 2x2- 2 = 5 - 2(x2 + 2x + 1) = 5 - 2(x+1)2  5 víi mäi x (2) Tõ (1) vµ (2) suy ra ph¬ng tr×nh cã nghiÖm  x = -1 Thö x = -1  lµ nghiÖm cña (*) Bµi 2:. Gi¶i ph¬ng tr×nh: 1 x  2  y  3  z  5  ( x  y  z  7) 2 . 2. . 2.  .  . x 2  1 . y 3 1 . . 2. z  5  1 0.  x 3    y 4  z 6 . Bµi 3:. Gi¶i ph¬ng tr×nh:. x y  1  2y x  1 .  y 1  §K  x 1. 3 xy 2.  2 x y  1  4 y x  1 3xy  xy  2 x y  1  2 xy  4 y x  1 0.   x. . . .  x y  2 y  1  2 y x  2 x  1 0. . 2. x 1  x. . y  1  1 2y. . . 2. x  1  1 0. (1) Do. . 2. y  1  1 0. y 1  2 y. . . dÊu “=” x¶y ra khi y = 2. 2. x  1  1 0. dÊu “=” x¶y ra khi x = 2 VËy nghiÖm cña ph¬ng tr×nh (1) lµ: x = 2 y=2 Bµi 4: 2 x 2  10 x  13  26 x 2  24 x  8 4 x  1 2.  4x  4  x2  6x  9 . . x. .  x  2.  . 2. . 2.   x  3 .  x  2. 2. x. 2.  4 x  4  25 x 2  20 x  4 4 x  1.  . 2.   5 x  2  4 x  1. .

<span class='text_page_counter'>(2)</span>  x  2. 2.   x  3  x  3 3  x. 2.  x  2. 2.   5 x  2   5 x  2 5 x  2. 2. Ta cã:  VT 5 x  2  3  x 4 x 1 DÊu “=” x¶y ra DÊu “=” x¶y ra DÊu “=” x¶y ra 3  x 0   5  x 0  x 2  x  2 0  2 VËy S =  . Bµi 5: a, Gi¶i hÖ ph¬ng tr×nh:.  x  1  y  1 4   x  y  xy 3.  xy 0 x  0 x  y  xy  3  0    y  0 §K:  x 1 0, y 1 0 mµ 2 x  2 y  xy  4 x  1  4 y  1  16  6 0.  x  y  2 xy    x 1  4 x 1  4   y 1  4  x  y    x 1  2    y 1  2  0 2. 2. 2.  x y    x  1 2  x  y 3   y  1 2. b, Gi¶i hÖ ph¬ng tr×nh: 4 xy 1  §K:  xy 0. 1  xy   4.  z 2  1 2 xy  2  x  1 2 yz 1  4 xy xy . 1 2 xy  1  “=” xÈy ra  xy = 4 z 2  1 1 “=” xÈy ra  z = 0  x 1  x  1   1 1   y   y  4 4   z o z 0 z = 0   hoÆc . 1 2. . y  1  4 0.

<span class='text_page_counter'>(3)</span> Bµi 6:. Gi¶i hÖ ph¬ng tr×nh:  x 2  xy  y 2 3  2  z  yz  1 0. 2   y y2   2 3y2 2 x   3 1        x  xy  y 3  2 4    4   2 2 2 y y y y2  z 2  xy    1  z  2   4  1 4 4     y2 0 1  y2 4   2  1  y 2 4  y  1 0  4.  S   1; 2;  1 ;   1;  2;1 . Bµi 7:. Gi¶i ph¬ng tr×nh:.  x2  4x  2   2x2  8x  5  2  3. . 2   x2  4 x  4  3   2x2  8x  8  2  3. . 2   x  2  3  2  x  2  2  3. 2.  2 x 2 2  2       3  2  x  2 2  3 . 2. " "  x 2 " "  x 2. 2. 2. 2   x  2  3  2  x  2  2  3. .  x 2  S  2. Bµi 8:. Gi¶i ph¬ng tr×nh:. a, x  1  5 x  1  3 x  2  x  5x.  x  1  5x  1. . x  1  5x  1. . x 1. Mµ. §K : x 1. 5x  1  0. 3 x  2  0  pt cã S   x 3x  2  2 x 3x  2. b, Gi¶i: . x 3x  2  2 x 3x  2. DK : x . 2 3.

<span class='text_page_counter'>(4)</span> x 3x  2  x 3x  2. . DÊu “=’.  x 2 3 x  2.  x 1  x 2  3x  2 0    x 2.  S  1; 2. Bµi 9:. Gi¶i x  1 2  x  1 . DK : x 1. x 1  x  1.  2  x 1  x  1  S   1 1 1  x  y  z 2    2  1 4 2 Gi¶i :  xy z. Bµi 10:. 2. 1 1  1      2   Tõ (1)  x y   z  1 1 4  2  2  0 x y z  1 1  2  4 2  2 x y   1 4     2   4  x x  . . 2. 1   x.  1 2    y 1  x y  2. (1). 2. . 1 1 2 1 4  2 4  2  2 x y xy z z. 1 1   0 y x  1 4   4  0 2 y y  2.  2  0   z. 1 2. 2, ¸p dông B§T C« si:. Bµi 1:. x 2  x  1   x 2  x  1 x 2  x  2 2. . 1 7  x  x  1   x  x  1  x    2 4  2. Ta cã §K:. 2. 2  x  x  1 0  2   x  x  1 0. Khi đó áp dụng: ta cã:. a. a 1 2. " " khi a 1. x2  x  1   x2  x 1 . x2  x  1  x2  x 1  2 2.

<span class='text_page_counter'>(5)</span> . x 2  x  1   x2  x  1 x  1. MÆt kh¸c: x 2  x  2  x  1  x 2  2 x  1 2.  x  1   x  1  x  1. VËy x 2  x  1   x 2  x  1 x 2  x  2 x  1  x 2  x  1 1    x 2  x  1 1  2 1  x  1.  x 1. VËy x=1 lµ nghiÖm Bµi 2: x2 x   1  2 x3  x 2  x  1 2 2 x2 x    1  ( x 2  x  1)(2 x  1) 2 2. (1). Ta cã x2 - x + 1 > 0 víi mäi x suy ra §K ¸p dông C«si cho 2 sè x2 – x + 1 > 0 2x + 1 > 0 ( x 2  x  1)(2 x  1) . x. 1 2. x2  x 1  2x 1 x2 x   1 2 2 2. Ta cã: VËy dÊu “=” x¶y ra  x2 – x + 1 = 2x +1  x2 – 3x = 0  x = 0 hoÆc x = 3. TM TM. 0;3 VËy S =  . 1 2 3    12 x y z  x  2 y  3 z 3 . Bµi 3: Gi¶i hÖ ph¬ng tr×nh: Víi x, y, z > 0. (1). 1 2 3    x  y  z 6 Tõ (1) ta cã: 4 x 4 y 4 z. V× x, y, z > 0 ta ¸p dông B§T C«si cho 2 sè (1). 1  x 1 4x. (2).  1  2  2 y 2   y  2 4x  4y . dÊu “=” x¶y ra khi dÊu “=” x¶y ra khi. x. 1 2. y. 1 2. 3 1   1 3  z   3 z 4z 4z   2 (3) dÊu “=” x¶y ra khi 1 2 3 x   2y   3z  1  2  3 6 4x 4y 4z Tõ (1), (2) vµ (3) ta cã: 3z .

<span class='text_page_counter'>(6)</span> dÊu “=” x¶y ra khi. x  y z . 1 2. TM.  1 1 1    , ,   S =  2 2 2  . vËy nghiÖm cña hÖ ph¬ng tr×nh lµ: Bµi 4: Gi¶i ph¬ng tr×nh: 2007 x2008 – 2008 x2007 + 1 = 0  1 + 2007 x2008 = 2008 x2007. ¸p dông B§T C«si cho 2008 sè d¬ng 1; x2008 ; x2008; x2008 …; x2008 ( 2007 sè x2008 ) 2008.  x>0. 2008 2007. 1.( x ) Ta cã: x + x + … + 1  2008 = 2008. x2007 dÊu “=” x¶y ra khi chØ khi 1 = x2008  x = 1 v× x > 0 VËy ph¬ng tr×nh cã nghiÖm x = 1 2008. 2008. 4. Bµi 5: Gi¶i ph¬ng tr×nh: x3 – x2 – 8x + 40 = 8 4 x  4 §K 4x + 4  0  x  -1 Víi § K x  -1 ta ¸p dông B§T C«si cho bèn sè: 4; 4; 4; x+1 ta cã: 4. 4 + 4 + 4 + x + 1 4. 4.4.4.( x  1). 4. =8. 4( x  1). 4.  13 + x  8 4( x  1)  13 + x  x3 – 3 x2 – 8x + 40  x3 – 3 x2 – 9 x + 27  0  ( x – 3 )2( x + 3 )  0 Do x  - 1  x + 3 > 0  ( x – 3 )2  0  x = 3 TM. VËy x = 3 lµ nghiÖm cña ph¬ng tr×nh. 2. Bµi 6: Gi¶i ph¬ng tr×nh: 7  x  x  5 x  12 x  38 (1) § K 5 x 7 Khi đó áp dụng BĐT áp dụng BĐT Côsi cho hai số 7  x 1 2 7 – x vµ 1 ta cã: x  5 1 x 5  2 x – 5 vµ 1 ta cã: 7  x 1 x  5 1 7 x  x 5   2  2 2 7 x . dÊu “=” x¶y ra khi chØ khi 7 – x = 1 x–5=1.  x=6. Ta l¹i cã: x2 – 12x + 38 = ( x – 6 )2 + 2  2 dÊu “=” x¶y ra khi chØ khi x = 6 6. VËy S =   Bµi tËp t¬ng tù: Bµi 1: Gi¶i ph¬ng tr×nh:. x 2  x  1   x 2  x 1 x 2  x  2. Bµi 2: Gi¶i ph¬ng tr×nh:. 2 x  3  5  2 x 3 x 2  12 x  14. Bµi 1: Gi¶i ph¬ng tr×nh:. 3, bất đẳng thức Bunhiacốpxki. 2 x  3  5  2 x 3 x 2  12 x  14 . 2. 2 x  3  5  2 x 3  x  2   2.

<span class='text_page_counter'>(7)</span> 2 x  3 0  1,5  x 2,5  5  2 x  0  §K:. ¸p dông Bu nhi a cèp xki cho (1:1) vµ ( 2 x  3 : 5  2x ). . 2x  3  5  2x. 2. 2 5  2 x  2.2 4  Do 2 x  3  5  2 x  0.   1 1    2. 2. 2 x  3  5  2 x 2. 2x  3. 2.  . DÊu “=” x¶y ra  2 x  3  5  2 x. .  x 2. 2. 3  x  2  2  2. dÊu”=” xÈy ra  x = 2 VËy pt cã nghiÖm duy nhÊt x = 2 Bµi 2: Gi¶i ph¬ng tr×nh a,. 6 2. A  x  2  5  2x . §K:. 2 x .  1. 5 2. 2    5 A  x  2.1   x . 2   2    Ta cã : 6 A 0  A  2 5 x  2. 2   x 1   2  xÈy ra  2. 13  S   6 b, 2 x  1  3 5  x 2 13. 2. x  1 3 5  x. 2.   2. 2. . x 2. . 2. 2  5   2 1    x   1  22  . 3 2  2  . . 13  x 6. . (TM§K). DK : 1 x 5.  32.   x  1  5  x  13.4. 2 x  1  3 5  x  2 13. PT x¶y ra.  3 x  1 2 5  x. 29  x  TM 13  29   S    13 . c, x 2  4 x  5 2 2 x  3 . . . 2. 2. 2 x  3  1   x  1 0.  x  1. Bµi 3: Gi¶i ph¬ng tr×nh :. x  2  10  x x 2  12 x  40. 2. x 2  12 x  10  x  6   4 4. DÊu “=” x¶y ra khi x = 6 Ta cã. . x  2  10  x. . 2. . .  x  2  10  x  12  12 16. DK :2 x 10.

<span class='text_page_counter'>(8)</span> x  2  10  x 4. D©u. “=”. Do : x  2  10  x  0. xÈy ra. x = 6 (TM).  S  6 x  1  x  3  2( x  3) 2  2 x  2. Bµi 4: Gi¶i ph¬ng tr×nh :. (1). ¸p dông B§T Bunhiac«pxki cho x  1 ; x – 3 vµ 1 ; 1 ta cã:.  . 2. 2 x  1  x  3  12  12    x  1   x  3   .  x  1  x  3 . 2( x  1)  2( x  3) 2. (2) x  1 x  3. (1) vµ (2) x¶y ra khi chØ khi:.  x2 – 6x + 9 = x – 1  x2 – 7x + 10 = 0  x=2. hoÆc x = 5. x = 2 kh«ng tho¶ m·n; x = 5 tho¶ m·n S  5. vËy. x2. Bµi 5: Gi¶i ph¬ng tr×nh :  x2. 4. 4. 2  x 4  1 x 4  x3. 2  x 4  1  x 3 ( x  1). § K : x4  2 x2 (. 4. 4. . 2  x 4  1 1  x 4. (x 0). 1  x2 2 x. 2  x4  x . 1  x 2 2 2 Ta cã: x. dÊu “=” x¶y ra 2.  4 2  x 4   12  12      MÆt kh¸c: . . 4. 2  x4  x 4.  . . 4. 4. 2  x2  x . 4. . .  x2 . 1 x2. 2  x4  x2. 2  x4  x2. . 2.  x 2 1. (1). . 4.2  2  x 4  x 4  16. 16 2. DÊu “=” x¶y ra khi chØ khi x = 1 Tõ (1) vµ (2) suy ra ph¬ng tr×nh cã nghiÖm cña nã lµ 1 TM 1. VËy S =   Bµi tËp t¬ng tù: Bµi tËp 1: Gi¶i ph¬ng tr×nh: Bµi tËp 2:. 6x  3 3  2 x  x 2 x  1 x. 6 x 2  3 xy  x 1  y  2 2 Gi¶i hÖ ph¬ng tr×nh:  x  y 1. (2).

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