Question

[4/(x^2-4)]+[1/(x+2)]/[1/(x-2)]

Answer 1

4/(x²-4)+[1/(x+2)]/[1/(x-2)]
=4/[(x+2)(x-2)]+[1/(x+2)](x-2)
=4/[(x+2)(x-2)]+(x-2)/(x+2)
=[4+(x-2)²]/[(x+2)(x-2)]
=(4+x²-4x+4)/(x²-4)
=(x²-4x+8)/(x²-4)

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追问

[(x-1)/(x^+x)]/[(x^2-2x+1)/(x^2-1)]-(1/x)

追答

[(x-1)/(x²+x)]/[(x²-2x+1)/(x²-1)]-(1/x)
=[(x-1)/x(x+1)]/[(x-1)²/(x+1)(x-1)]-(1/x)
=[(x-1)/x(x+1)][(x+1)(x-1)/(x-1)²]-(1/x)
=(1/x)-(1/x)
=0

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