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已知x+2=1/x,试求代数式1/(x+1)-(x+3)/(x的平方-1)×(x的平方-2x+1)/(x的平方+4x+3)的值.
分式方程.
人气:252 ℃ 时间:2019-08-19 01:51:08
解答
x+2=1/xx²+2x=11/(x+1)-(x+3)/(x的平方-1)×(x的平方-2x+1)/(x的平方+4x+3)=1/(x+1)-(x+3)/(x+1)(x-1) x (x-1)²/(x+1)(x+3) =1/(x+1)-(x-1)/(x+1) ²=(x+1-x+1)/(x+1)²=2/(x+1)²=2/(x&#...
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