Toán Lớp 8: a.(x ²-2) ²+4(x-1) ²-4(x ²-2)(x-1)=0 b.(x+2) ²=(2x-1) ²

Question

Toán Lớp 8:  a.(x ²-2) ²+4(x-1) ²-4(x ²-2)(x-1)=0 b.(x+2) ²=(2x-1) ²

hiennhi98 · Answer

Lời giải và giải thích chi tiết:    a)   ( 2x+1) &sup2; -4x ( x-1) =5     &hArr; (2x)&sup2; + 2.2x + 1 - (4x.x -4x.1) = 5     &hArr; 4x&sup2; + 4x + 1 - 4x&sup2; +4x = 5     &hArr; 8x = 4 &hArr;x = 1/2     b)  ( x-2)(x+2)+(x+2) &sup2; = 2x(x-4)     &hArr; (x&sup2; - 4) +(x&sup2; +2.2x +2&sup2;) = 2x.x -4.2x       &hArr; x&sup2; - 4 + x&sup2; + 4x + 4 = 2x&sup2; - 8x     &hArr; 4x +8x = 0     &hArr; x = 0     c) (3x-2) &sup2;-(3x+2) &sup2;=8      &hArr; (3x-2 + 3x+2)(3x-2 -3x - 2) = 8 (N&agrave;y l&agrave; AD HĐT lu&ocirc;n th&ocirc;i ạ)     &hArr; 6x(-4) = 8      &hArr; 6x = -2     &hArr; x = -1/3

cobelolen · Answer

a)   (x^2 - 2)^2 + 4 (x-1)^2 - 4 (x^2 - 2)(x-1) = 0   => (x^2-2)^2  + [2 (x-1)]^2 - 2  (x-1) . (x^2 - 2) =0   => [ (x^2 - 2) - 2 (x-1)]^2 =0   => (x^2 - 2 - 2x  +2)^2=0   => (x^2-2x)^2=0   =>x^2-2x=0   => x (x-2) =0   =>x=0 hoặc x-2=0   +)x=0   +)x-2=0=>x=2   Vậy x  in {0;2}   b)   (x+2)^2 = (2x-1)^2   => (x+2)^2-(2x-1)^2=0   => [ (x+2) - (2x-1)] [ (x+2) + (2x-1)]=0   => (x+2-2x+1)(x+2+2x-1)=0   => (3-x)(3x+1)=0   =>3-x=0 hoặc 3x+1=0   +)3-x=0=>x=3   +)3x+1=0=>3x=-1=>x=-1/3   Vậy x in{3;-1/3}