Question

初二数学，就两题，急急急，过程

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Answer 1

（1）1/(a-b) - (a+b)/(a-b)^2 - (a+2b)/(a^2-b^2)
= (a^2-b^2 - (a+b)^2 - (a+2b)(a-b)))/(a+b)(a-b)^2
= (a^2-b^2 -a^2-b^2 -2ab - a^2 - 2ab + ab + 2b^2) / (a+b)(a-b)^2
= (-a^2 - 3ab ) / (a+b)(a-b)^2
= -a(a+3b) / (a+b)(a-b)^2

（2）1/(x-1) + 1/(x-2) - (x+3)/(x^2+x-2)
= 1/(x-1) + 1/(x-2) - (x+3)/(x+2)(x-1)
= (x^2-4 + (x+2)(x-1) - (x+3)(x-2)) / (x+2)(x-1)(x-2)
= (x^2-4 + x^2+x-2 - x^2-x+6) / (x+2)(x-1)(x-2)
= x^2 / (x+2)(x-1)(x-2)

答题不易，请及时采纳，谢谢！