Ap dung bat dang thuc de giai PT He PT

3, bất đẳng thức Bunhiacốpxki.[r]

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: x  x  12  x  10 x  3  x  x (*) Gi¶i: Ta cã: 3x + 6x + 12 = 3x + 6x + + = 3(x +1)2 +  víi mäi x 5x2+ 10x + = 5x2+ 10x + + = 5(x + 1)2+  víi mäi x 2 x  x  12  x  10 x    5 (1)  Mµ - 4x - 2x2 = - 4x- 2x2- = - 2(x2 + 2x + 1) = - 2(x+1)2  víi mäi x (2) Tõ (1) vµ (2) suy ph¬ng tr×nh cã nghiÖm  x = -1 Thö x = -1  lµ nghiÖm cña (*) Bµi 2: Gi¶i ph¬ng tr×nh: x   y   z   ( x  y  z  7)       x   y 3   z   0  x 3    y 4  z 6  Bµi 3: Gi¶i ph¬ng tr×nh: x y   2y x    y 1  §K  x 1 xy  x y   y x  3xy  xy  x y   xy  y x  0   x     x y  y   y x  x  0  x 1  x  y   2y   x   0 (1) Do  y   0 y 1  y   dÊu “=” x¶y y = 2 x   0 dÊu “=” x¶y x = VËy nghiÖm cña ph¬ng tr×nh (1) lµ: x = y=2 Bµi 4: x  10 x  13  26 x  24 x  4 x   4x   x2  6x    x   x  2      x  3   x  2 x  x   25 x  20 x  4 x      x   4 x    x  2   x  3  x  3  x  x  2   x    x  5 x  2 Ta cã:  VT 5 x    x 4 x 1 DÊu “=” x¶y DÊu “=” x¶y DÊu “=” x¶y 3  x 0   5  x 0  x 2  x  0  VËy S =   Bµi 5: a, Gi¶i hÖ ph¬ng tr×nh:  x   y  4   x  y  xy 3  xy 0 x  x  y  xy      y  §K:  x 1 0, y 1 0 mµ x  y  xy  x   y   16  0  x  y  xy    x 1  x 1  4   y 1   x  y    x 1     y 1   0 2  x y    x  2  x  y 3   y  2 b, Gi¶i hÖ ph¬ng tr×nh: 4 xy 1  §K:  xy 0  xy    z  2 xy   x  2 yz  xy xy  xy   “=” xÈy  xy = z  1 “=” xÈy  z =  x 1  x    1   y   y  4   z o z 0 z =   hoÆc   y   0 Bµi 6: Gi¶i hÖ ph¬ng tr×nh:  x  xy  y 3   z  yz  0   y y2   3y2 x          x  xy  y 3  2      2 y y y y2  z  xy    1  z     4     y2 0 1  y2    1  y 2  y  0   S   1; 2;  1 ;   1;  2;1  Bµi 7: Gi¶i ph¬ng tr×nh:  x2  4x    2x2  8x       x2  x  4    2x2  8x  8      x  2    x  2    2 x 2          x  2   " "  x 2 " "  x 2 2   x  2    x  2     x 2  S  2 Bµi 8: Gi¶i ph¬ng tr×nh: a, x   x   x   x  5x  x   5x   x   5x   x 1 Mµ §K : x 1 5x   x    pt cã S   x 3x   2 x 3x  b, Gi¶i:  x 3x   2 x 3x  DK : x  x 3x   x 3x   DÊu “=’  x 3 x   x 1  x  3x  0    x 2  S  1; 2 Bµi 9: Gi¶i x  2  x   DK : x 1 x 1  x    x 1  x   S   1 1  x  y  z 2     4 Gi¶i :  xy z Bµi 10: 1 1  1        Tõ (1)  x y   z  1    0 x y z  1   4  x y         4  x x    1   x  1 2    y  x y  (1)  1  2 4   x y xy z z 1   0 y x     0 y y    0   z 2, ¸p dông B§T C« si: Bµi 1: x  x    x  x  x  x  2  1  x  x    x  x   x    2  Ta cã §K: 2  x  x  0    x  x  0 Khi đó áp dụng: ta cã: a a 1 " " a 1 x2  x    x2  x 1  x2  x   x2  x 1  2  x  x    x2  x  x  MÆt kh¸c: x  x   x  1  x  x   x  1   x  1  x  VËy x  x    x  x  x  x  x   x  x  1    x  x  1  1  x  1  x 1 VËy x=1 lµ nghiÖm Bµi 2: x2 x    x3  x  x  2 x2 x     ( x  x  1)(2 x  1) 2 (1) Ta cã x2 - x + > víi mäi x suy §K ¸p dông C«si cho sè x2 – x + > 2x + > ( x  x  1)(2 x  1)  x 1 x2  x 1  2x 1 x2 x   1 2 Ta cã: VËy dÊu “=” x¶y  x2 – x + = 2x +1  x2 – 3x =  x = hoÆc x = TM TM 0;3 VËy S =   1    12 x y z  x  y  z 3  Bµi 3: Gi¶i hÖ ph¬ng tr×nh: Víi x, y, z > (1)    x  y  z 6 Tõ (1) ta cã: x y z V× x, y, z > ta ¸p dông B§T C«si cho sè (1)  x 1 4x (2)    y 2   y  2 4x  4y  dÊu “=” x¶y dÊu “=” x¶y x y   3  z   3 z 4z 4z   (3) dÊu “=” x¶y x   2y   3z  1   6 4x 4y 4z Tõ (1), (2) vµ (3) ta cã: 3z  dÊu “=” x¶y x  y z  TM  1    , ,   S =  2   vËy nghiÖm cña hÖ ph¬ng tr×nh lµ: Bµi 4: Gi¶i ph¬ng tr×nh: 2007 x2008 – 2008 x2007 + =  + 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 + … +  2008 = 2008 x2007 dÊu “=” x¶y chØ = x2008  x = v× x > VËy ph¬ng tr×nh cã nghiÖm x = 2008 2008 Bµi 5: Gi¶i ph¬ng tr×nh: x3 – x2 – 8x + 40 = x  §K 4x +   x  -1 Víi § K x  -1 ta ¸p dông B§T C«si cho bèn sè: 4; 4; 4; x+1 ta cã: 4 + + + x + 4 4.4.4.( x  1) =8 4( x  1)  13 + x  4( x  1)  13 + x  x3 – x2 – 8x + 40  x3 – x2 – x + 27   ( x – )2( x + )  Do x  -  x + >  ( x – )2   x = TM VËy x = lµ nghiÖm cña ph¬ng tr×nh Bµi 6: Gi¶i ph¬ng tr×nh:  x  x  x  12 x  38 (1) § K x 7 Khi đó áp dụng BĐT áp dụng BĐT Côsi cho hai số  x 1 – x vµ ta cã: x  1 x  x – vµ ta cã:  x 1 x  1 7 x  x   2  2 7 x  dÊu “=” x¶y chØ – x = x–5=1  x=6 Ta l¹i cã: x2 – 12x + 38 = ( x – )2 +  dÊu “=” x¶y chØ x = 6 VËy S =   Bµi tËp t¬ng tù: Bµi 1: Gi¶i ph¬ng tr×nh: x  x    x  x 1 x  x  Bµi 2: Gi¶i ph¬ng tr×nh: x    x 3 x  12 x  14 Bµi 1: Gi¶i ph¬ng tr×nh: 3, bất đẳng thức Bunhiacốpxki x    x 3 x  12 x  14  2 x    x 3  x    2 x  0  1,5  x 2,5   x   §K: ¸p dông Bu nhi a cèp xki cho (1:1) vµ ( x  :  2x )  2x    2x 2  x  2.2 4  Do x    x    1    2 x    x 2 2x    DÊu “=” x¶y  x    x   x 2  x  2   dÊu”=” xÈy  x = VËy pt cã nghiÖm nhÊt x = Bµi 2: Gi¶i ph¬ng tr×nh a, A  x    2x  §K: x   1 2    A  x  2.1   x      Ta cã : A 0  A  x  2   x    xÈy  13  S   6 b, x    x 2 13 2 x  3  x   2  x  2       x    22      13  x  (TM§K) DK : x 5  32   x    x  13.4 x    x  13 PT x¶y  x  2  x 29  x  TM 13  29   S    13  c, x  x  2 x     2 x     x  1 0  x  Bµi 3: Gi¶i ph¬ng tr×nh : x   10  x x  12 x  40 x  12 x  10  x    4 DÊu “=” x¶y x = Ta cã  x   10  x     x   10  x  12  12 16 DK :2 x 10 x   10  x 4 D©u “=” Do : x   10  x  xÈy x = (TM)  S  6 x   x   2( x  3)  x  Bµi 4: Gi¶i ph¬ng tr×nh : (1) ¸p dông B§T Bunhiac«pxki cho x  ; x – vµ ; ta cã:   2 x   x   12  12    x  1   x  3     x   x  3  2( x  1)  2( x  3) (2) x  x  (1) vµ (2) x¶y chØ khi:  x2 – 6x + = x –  x2 – 7x + 10 =  x=2 hoÆc x = x = kh«ng tho¶ m·n; x = tho¶ m·n S  5 vËy x2 Bµi 5: Gi¶i ph¬ng tr×nh :  x2 4  x  x  x3  x   x ( x  1) § K : x4  x2 ( 4   x  1  x (x 0)  x2 x  x4  x   x 2 Ta cã: x dÊu “=” x¶y   x   12  12      MÆt kh¸c:    x4  x    4  x2  x     x2  x2  x4  x2  x4  x2   x 1 (1)  4.2   x  x  16 16 2 DÊu “=” x¶y chØ x = Tõ (1) vµ (2) suy ph¬ng tr×nh cã nghiÖm cña nã lµ TM VËy S =   Bµi tËp t¬ng tù: Bµi tËp 1: Gi¶i ph¬ng tr×nh: Bµi tËp 2: 6x  3  x  x x  1 x 6 x  xy  x 1  y  2 Gi¶i hÖ ph¬ng tr×nh:  x  y 1 (2)