用因式分解法解解一元二次方程〔X+1〕²-2(X+1)(2-X)+(2-X)²=0

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用因式分解法解解一元二次方程〔X+1〕²-2(X+1)(2-X)+(2-X)²=0
用因式分解法解
解一元二次方程〔X+1〕²-2(X+1)(2-X)+(2-X)²=0

用因式分解法解解一元二次方程〔X+1〕²-2(X+1)(2-X)+(2-X)²=0
〔X+1〕²-2(X+1)(2-X)+(2-X)²=0
1 1
1 1
就是把(x+1)和(2-x)看成x和y即可
[(x+1)-(2-x)]²=0
(2x-1)²=0
2x-1=0
x1=x2=1/2

〔X+1〕²-2(X+1)(2-X)+(2-X)²=0
〔X+1〕²+2(X+1)(X-2)+(X-2)²=0
(X+1+X-2)^2=0
2X-1=0
X=1/2


(X+1〕²-2(X+1)(2-X)+(2-X)²=0
[x+1-(2-x)]²=0
(x+1-2+x)²=0
(2x-1)²=0
2x-1=0
x=1/2

原式=[(x+1)-(2-x)]^2=(2x-1)^2=0
x=1/2

设x+1=t ;2-x=m
则原式可写成t^2-2tm+m^2=(t-m)^2=0
代入得:(x+1-2+x)^2=0
即2x-1=0,x=1/2

(X+1)(X+1-(2-X))+(2-X)(2-X-(X+1))=0
(X+1)(2X-1)+(2-X)(1-2X)=0
(2X-1)(X+1-(2-X))=0
(2X-1)(2X-1)=0
(2X-1)^2=0
2x-1=0
...

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(X+1)(X+1-(2-X))+(2-X)(2-X-(X+1))=0
(X+1)(2X-1)+(2-X)(1-2X)=0
(2X-1)(X+1-(2-X))=0
(2X-1)(2X-1)=0
(2X-1)^2=0
2x-1=0
X=1/2 (0.5)
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把2(x+1)(2-x)拆开(x+1)(2-x)+(x+1)(2-x)再因式分解