Toán Lớp 8: tìm x a.x^2 -2x-15=0 b.(2x+1)^2 -4(x-1)(x+1)=9 c. (x+2)(x^2-2x+4)-x(x-1)^2=2x^2

Question

Toán Lớp 8:  tìm x  a.x^2 -2x-15=0 b.(2x+1)^2 -4(x-1)(x+1)=9 c. (x+2)(x^2-2x+4)-x(x-1)^2=2x^2

quynhmaithu · Answer

$ text{a) x&sup2; - 2x - 15 = 0}$   $ text{&hArr;x&sup2; + 3x - 5x - 15 = 0}$   $ text{&hArr; x( x + 3 ) - 5( x + 3 ) = 0}$   $ text{&hArr; ( x - 5 )( x + 3 ) = 0}$   &hArr; $ left[ begin{matrix} x =5   x = -3 end{matrix} right.$   $ text{b) ( 2x + 1 )&sup2; - 4( x + 1 )( x - 1 ) = 9}$   $ text{&hArr; 4x&sup2; + 4x + 1 - 4( x&sup2; - 1 ) = 9}$   $ text{&hArr; 4x&sup2; + 4x + 1 - 4x&sup2; + 4 = 9}$   $ text{&hArr; ( 4x&sup2; - 4x&sup2; )  + 4x = 9 - 4 - 1}$   $ text{&hArr; 4x = 4}$   $ text{&hArr; x = 1}$   $ text{c) ( x + 2 )( x&sup2; - 2x + 4 ) - x( x - 1 )&sup2; = 2x&sup2;}$   $ text{&hArr; x&sup3; + 8 - x ( x&sup2; - 2x + 1 ) = 2x&sup2;}$   $ text{&hArr; x&sup3; + 8 - x&sup3; + 2x&sup2; - x = 2x&sup2;}$   $ text{&hArr; ( x&sup3; - x&sup3; ) + 8 -x = 2x&sup2; - 2x&sup2;}$   $ text{&hArr; 8 - x = 0}$   $ text{&hArr; x = 8}$

ngocbichchau · Answer

a)  $ x^2 - 2x - 15 =0 $   &rArr; $ x^2 + 3x - 5x - 15 = 0 $   &rArr; $ x( x + 3 ) - 5 ( x + 3 ) =  0 $   &rArr; $ ( x+ 3 ) ( x - 5 ) = 0 $   &rArr;  ( left[  begin{array}{l}x+3=0  x-5=0 end{array}  right. )    &rArr;  ( left[  begin{array}{l}x=-3  x=5 end{array}  right. )    Vậy $ x = - 3 $ hoặc $ x = 5 $    b)  $ ( 2x + 1)^2 - 4 ( x- 1 ) ( x + 1 ) = 9 $   &rArr;$ ( 2x)^2 + 2 . 2x . 1 + 1^2 - 4 ( x^2 - 1 ) - 9 = 0 $   &rArr;$ 4x^2 + 4x + 1 - 4x^2 + 4 - 9 = 0 $   &rArr; $4x -4 = 0 $   &rArr; $4x = 4 $   &rArr;$ x = 1 $   Vậy $ x = 1 $    c)  $ ( x + 2 ) ( x^2 - 2x + 1 ) - x ( x - 1 )^2 = 2x^2$   &rArr; $ x^3 + 2^3 - x ( x^2 - 2x + 1 ) - 2x^2 = 0 $   &rArr; $ x^3 + 8 - x^3 + 2x^2 - x - 2x^2 = 0 $   &rArr; $ 8 - x = 0 $   &rArr; $ x = 8 $   Vậy $ x = 8 $