Question

先化简,在求值1+x/x²+x-2÷(x-2+3/x+2),其中x=1/2

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Answer 1

原式={（1+x）/ [ (x+2)(x-1) ] }乘以{ (x+2) / [(x-1)(x+1)] }=1/[(x-1)(x-1)]=4

Answer 2

(1+x)/(x²+x-2)÷[(x-2)+3/(x+2)]
=(x+1)/(x+2)(x-1)÷{[(x-2)(x+2)+3]/(x+2)}
=(x+1)/(x+2)(x-1)*(x+2)/(x-1)(x+1)
=1/(x-1)^2
=1/(1/2-1)^2 (其中x=1/2)
=1/(1/4)
=4