Toán Lớp 8: Bài 1 ( 2điểm ): Tìm x biết: a, x (x – 3) – x 2 + x = 6 b, 2x. (x – 1) = (x – 1) 2 c, x 3 + 4x 2 + 4x = 0

Question

Toán Lớp 8:  Bài 1 ( 2điểm ): Tìm x biết: a,  x (x – 3) – x 2 + x = 6                          b, 2x. (x – 1) = (x – 1) 2 c, x 3 + 4x 2 + 4x = 0                               d, x 2 + x – 2 = 0

lanngocha · Answer

Giải đáp:     a) x = - 3     b) x = 1 ; x = - 1     c) x = 0 ; x = - 2     d) x = -2 ; x = 1     Lời giải và giải thích chi tiết:    a)   x(x - 3) - x^2 + x = 6   x^2 - 3x - x^2 + x - 6 = 0   (x^2 - x^2) - (3x - x) - 6 = 0   - 2x - 6 = 0   - 2(x + 3) = 0   => x + 3 = 0   => x = - 3   Vậy x = - 3   -------------------------------------------   b)   2x(x - 1) = (x - 1)^2   2x(x - 1) - (x - 1)^2 = 0   (x - 1)[2x - (x - 1)] = 0   ( x- 1)(2x - x + 1) = 0   (x - 1)(x + 1) = 0   => [(x - 1 = 0),(x + 1 = 0):}   => [(x = 1),(x = - 1):}   Vậy x = 1 ; x = - 1   -------------------------------------------   c)   x^3 + 4x^2 + 4x = 0   x(x^2 + 4x + 4) = 0   x(x^2 + 2.x.2 + 2^2) = 0   x(x + 2)^2 = 0   => [(x = 0),((x + 2)^2 = 0):}   => [(x = 0),(x + 2 = 0):}   =>[(x = 0),(x = - 2):}   Vậy x = 0 ; x = - 2   -------------------------------------------   d)   x^2 + x - 2 = 0   x^2 + 2x - x - 2  = 0   (x^2 + 2x) - (x + 2) = 0   x(x + 2) - (x + 2) = 0   (x + 2)(x - 1) = 0   => [(x + 2 = 0),(x - 1 = 0):}   => [(x = - 2),(x = 1):}   Vậy x = -2 ; x = 1   #SunHee

daithanh · Answer

Gửi cậu n&ograve; (❁&acute;◡❁)   a)   x.(x-3)-x^2+x=6   =>x.x-x.3-x^2+x=6   =>x^2-3x-x^2+x=6   =>-2x=6   =>x=6:(-2)   =>x=-3   Vậy x=-3   $  $   b)   2x.(x-1)=(x-1)^2   =>2x.(x-1)-(x-1)^2=0   =>(x-1).[2x-(x-1)]=0   =>(x-1).(2x-x+1)=0   =>(x-1).(x+1)=0   => ( left[  begin{array}{l}x-1=0  x+1=0 end{array}  right. )    => ( left[  begin{array}{l}x=1  x=-1 end{array}  right. )    Vậy x=1 v&agrave; x=-1   $  $   c)   x^3+4x^2+4x=0   =>x.(x^2+4x+4)=0   =>x.(x^2+2.x.2+2^2)=0   =>x.(x+2)^2=0   => ( left[  begin{array}{l}x=0  (x+2)^2=0 end{array}  right. )    => ( left[  begin{array}{l}x=0  x+2=0 end{array}  right. )    => ( left[  begin{array}{l}x=0  x=-2 end{array}  right. )    Vậy x=0 v&agrave; x=-2   $  $   d)   x^2+x-2=0   =>x^2+2x-x-2=0   =>(x^2+2x)-(x+2)=0   =>x.(x+2)-(x+2)=0   =>(x+2).(x-1)=0   => ( left[  begin{array}{l}x+2=0  x-1=0 end{array}  right. )    => ( left[  begin{array}{l}x=-2  x=1 end{array}  right. )    Vậy x=-2 v&agrave; x=1