Question

分解因式法解方程2（x-1）^2+x=1；？

Answer 1

2(x-1)² + x = 1
2(x² - 2x+1) +x - 1 = 0
2x² - 3x + 1 = 0
(x-1)(2x-1)=0
x₁ = 1 ,x₂ = 1/2,5,2(x-1)^2+x=1
2x^2-4x+2+x=1
2x^2-3x+1=0
(2x-1)(x-1)=0
x=1或x=1/2,2,2(x-1)²+x=1
∴2(x-1)²+(x-1)=0
∴(x-1)(2x-2+1)=0
即(x-1)(2x-1)=0
解得x=1/2或1.,2,先化开再进行因式分解
2（x2-2x+1)+x-1=o
2x2-4x+2+x-1=0
2x2-3x+1=0
(2x-1)(x-1)=o,1,解。原式等价于：2（x^2-2x+1）+x-1=0;
即 2*x^2-3x+1=0,
(2x-1)(x-1)=0;
解得，x=1或x=1/2,0,