Question

解方程2x^2-2根号2x+1 用因式分解法解

Answer 1

解;① (2x+3)^2=(5x-4)^2
(2x+3+5x-4)(2x+3-5x+4)=0 x1=1/5 x2=7/3
..② (x-根号下2）^2-2（根号下2-x)=0
(x-根号下2)(x-根号下2+2）=0 x1=根号2 x2=根号2-2
③ 4y^2+8y+4=0
4（y-1）^2=0 y1=y2=1
..④ (x-5)(x+3)+x(x+6)=-17
2x^2+4x+2=0 (x+1)^2=0 x1=x2=-1
..⑤ (3-x) ^2 =9-x ^2
(3-x) ^2-（3-x）（3+x）=0
(3-x)(3-x-3-x)=0 x1=3 x2=0
⑥(t-3)(t+4)=-12 t^2+t=0 t（t+2）=0 t1=0 t2=-2
..⑦ (x+5)(x+3)+x(x+6)=-17 2x^2+14x+32=0 x^2+7x+16=0 方程无解
..⑧x^2-2ax+a^2-b^2=0 [x-(a+b)][x-(a-b)]=0
x1=a+b x2=a-b