先化简再求值(x^2-1分之x^2-2x+1+x分之1)÷x+1分之1,其中x=-2
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解:/ 表示分数线
原式={[(x²-2x+1)/(x²-1)]+1/x}÷[1/(x+1)]
={[(x-1)²]/[(x+1)(x-1)]+1/x}×(x+1)
={[(x-1)/(x+1)]+1/x}×(x+1)
=[(x-1)/(x+1)]×(x+1)+[(x+1)/x]
=(x-1)+[(x+1)/x]
=[x(x-1)+(x+1)]/x
=(x²-x+x+1)/x
=(x²+1)/x
当x=-2时
原式=[(-2)²+1]/(-2)
=(4+1)/(-2)
= -5/2
原式={[(x²-2x+1)/(x²-1)]+1/x}÷[1/(x+1)]
={[(x-1)²]/[(x+1)(x-1)]+1/x}×(x+1)
={[(x-1)/(x+1)]+1/x}×(x+1)
=[(x-1)/(x+1)]×(x+1)+[(x+1)/x]
=(x-1)+[(x+1)/x]
=[x(x-1)+(x+1)]/x
=(x²-x+x+1)/x
=(x²+1)/x
当x=-2时
原式=[(-2)²+1]/(-2)
=(4+1)/(-2)
= -5/2
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