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用因式分解法解下列一元二次方程.(1)(x-8)2=5(x-8);(2)(X-8)2=(2X-3)2; (3)(X-5)2+2(X-5)+1=0
(4)9(X-1)2-4(X+2)2=0
人气:252 ℃ 时间:2019-11-04 08:36:29
解答
1.
(x-8)²=5(x-8)
(x-8)²-5(x-8)=0
(x-8)[(x-8)-5]=0
(x-8)(x-13)=0
x=8或x=13
2.
(x-8)²=(2x-3)²
(x-8)²-(2x-3)²=0
[(x-8)+(2x-3)][(x-8)-(2x-3)]=0
-(3x-11)(x+5)=0
x=11/3或x=-5
3.
(x-5)²+2(x-5)+1=0
[(x-5)+1]²=0
(x-4)²=0
x=4
4.
9(x-1)²-4(x+2)²=0
[3(x-1)²]-[2(x+2)]²=0
[3(x-1)+2(x+2)][3(x-1)-2(x+2)]=0
(5x+1)(x-7)=0
x=-1/5或x=7
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