Toán Lớp 8: 16. Tìm x, biết: a) $x^{2}+4=4x$ b) $4x^{2}-1=0$ c) $x^{3}-3x^{2}+3x=1$ d) $(x+1)^{2}-(2x-1)^{2}=0$

Question

Toán Lớp 8:  16. Tìm x, biết: a)  $x^{2}+4=4x$ b)   $4x^{2}-1=0$ c)   $x^{3}-3x^{2}+3x=1$  d)  $(x+1)^{2}-(2x-1)^{2}=0$

ngochuong2k2 · Answer

Giải đáp + Lời giải và giải thích chi tiết:    a)   x&sup2; + 4 = 4x   x&sup2; - 4x + 4 = 0   (x-2)&sup2; = 0   x = 2   b)   4x&sup2; - 1 = 0   (2x-1)(2x+1) = 0    ( left[  begin{array}{l}x= dfrac{1}{2}  x=- dfrac{1}{2} end{array}  right. )    c)   x&sup3; - 3x&sup2; + 3x = 1   x&sup3; - 3x&sup2; + 3x - 1 = 0   (x-1)&sup3; = 0    x = 1   d)   (x+1)&sup2; - (2x-1)&sup2; = 0   (x+1 - 2x + 1)(x+1 + 2x-1)= 0   (-x+2)3x = 0    ( left[  begin{array}{l}x=0  x=2 end{array}  right. )

ngoclankhue · Answer

a) x^2+4=4x   &hArr;x^2-4x+4=0   &hArr;(x-2)^2=0   &hArr;x-2=0   &hArr;x=2   Vậy S={2}   b) 4x^2-1=0   &hArr;(2x-1)(2x+1)=0   &hArr; ( left[  begin{array}{l}2x-1=0  2x+1=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x= dfrac{1}{2}  x=- dfrac{1}{2} end{array}  right. )    Vậy S={1/2,-1/2}   c) x^3-3x^2+3x=1   &hArr;x^3-3x^2+3x-1=0   &hArr;(x-1)^3=0   &hArr;x-1=0   &hArr;x=1   Vậy S={1}   d) (x+1)^2-(2x-1)^2=0   &hArr;(x+1-2x+1)(x+1+2x-1)=0   &hArr;(-x+2)3x=0   &hArr; ( left[  begin{array}{l}-x+2=0  3x=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=2  x=0 end{array}  right. )    Vậy S={2,0}