Question

用因式分解解下列方程: (x-1)^2-4(1-x)+4=0； x(2x＋1)=3＋6x；

用因式分解解下列方程:
(x-1)^2-4(1-x)+4=0；
x(2x＋1)=3＋6x；

Answer 1

(x-1)²-4(1-x)+4=0

解：(1-x)²-4(1-x)+4=0
另1-x=y 或直接（1-x-2）²=0
y²-4y+4=0 (-x-1)²=0
( y-2)²=0 x1=x2=-1
y1=y2=2
x1=x2=-1

Answer 2

解x²=2xx²-2x=0x(x-2)=0x=0或x=2(x+1)²-(2x-3)²=0(x+1+2x-3)(x+1-2x+3)=0(3x-2)(-x+4)=0∴x=2/3或x=4x²-6x+8=0(x-4)(x-2)=0∴x=4或x=24(x+3)³=25(x-2)²4(x+3)²-25(x-2)²=0[2(x+3)+5(x-2)][2(x+3)-5(x-2)]=0(7x-3)(-3x+16)=0∴x=3/7或x=16/3(1+√2)x²-(1-√2)x=0x[(1+√2)x-(1-√2)]=0∴x=0或x=(1-√2)/(1+√2)=(1-√2)²/(1-2)=-(1-2√2+2)=-3+2√2