Question

用分解因式法解一元二次方程，大家帮忙！！~~

用分解因式法解一元二次方程：[麻烦写过程，谢谢大家]
1. （2x-1）^2-x^2=0
2. x^2-6x+9=(5-2x)^2
3. 2(x+2)^2=x^2-4
4. 10(x-3)+x=x^2

Answer 1

1、平方差公式：（2x-1+x)(2x-1-x) = 0
(3x-1)(x-1) = 0
3x-1 = 0或x-1 = 0
解得x1=1/3，x2=1.
2、（x-3)²－(5-2x)² = 0
[(x-3)+(5-2x)][(x-3)－(5-2x)] = 0
(2-x)(3x-8) = 0
2-x= 0 或3x-8 = 0
x1=2，x2=8/3
3、2（x+2)² = (x+2)(x-2)
2（x+2)² － (x+2)(x-2) ＝ 0
(x+2)[2(x+2)-(x-2)] = 0
(x+2)(x+6) = 0
x1=-2, x2 = -6
4、整理得，x² －11x + 30 = 0
(x-5)(x-6) = 0
x1 = 5, x2 = 6.

Answer 2

1: （2x-1）^2-x^2=0
(2X-1+X)(2X-1-X)=0 (3X-1)(X-1)=0 X1=1/3 X2=1
2. x^2-6x+9=(5-2x)^2
( x-3)^2=(5-2x)^2 (x-3-5+2x)(x-3+5-2x)=0 (3x-8)(-x+2)=0 x1=8/3 x2=2
3. 2(x+2)^2=x^2-4
2(x+2)^2=(x+2)(x-2) 2(x+2)=x-2 x=-6
4. 10(x-3)+x=x^2
x^2-11x+30=0 (x-6)(x-5)=0 x1=5 x2=6

楼上的第三题做错了。。

Answer 3

1. （2x-1）^2-x^2=0
（2x-1-x)(2x-1+x)=0
(x-1)(3x-1)=0
x=1或x=1/3

2. x^2-6x+9=(5-2x)^2
(x-3)^2-(5-2x)^2=0
(x-3-5+2x)(x-3+5-2x)=0
(3x-8)(2-x)=0
x=8/3或x=2

3. 2(x+2)^2=x^2-4
2(x+2)^2-(x^2-4)=0
2(x+2)^2-(x-2)(x+2)=0
[2(x+2)-(x-2)](x+2)=0
(2x+4-x+2)(x+2)=0
(x+6)(x+2)=0
x=-6或x=-2

4. 10(x-3)+x=x^2
x^2-10(x-3)-x=0
x^2-10x+30-x=0
x^2-11x+30=0
(x-5)(x-6)=0
x=5或x=6

Answer 4

1. 4x^2-4x+1-x^2=0
3x^2-4x+1=0
（3x-1)(x-1)=0
2. x^2-6x+9=25-20x+4x^2
3x^2-16x+16=0
(3x-4)(x-4)=0
3. 2x^2+8x+8=x^2-4
x^2+8x+12=0
(x-2)(x-6)=0
4. 10x-30+x=x^2
x^2-11x+30=0
(x-5)(x-6)=0

Answer 5

1) (2x-1-x)*(2x-1+x)=0
(x-1)*(3x-1)=0
x=1或x=1/3