Question

初二几道分式计算1.b/(a-b)+a/(a+b)-2ab/(b^2-a^2)2.(2a+2)/(a+1)-(a^2-1

Answer 1

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

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

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

=(a-b)^2/(a+b)(a-b)

=a-b/a+b

2.=2(a+1)/(a+1)-(a^2-1^2)/(a-1)^2

=2-(a+1)(a-1)/(a-1)^2

=2-(a+1)/(a-1)

3 =x(x-2)/(x^2-1^2)-1/{(x+1)(x-1)/(x+1)++(2x-1)/(x+1)

=x(x-2)/(x^2-1^2)-1/(x^2-2x)/(x+1)

=x(x-2)/(x^2-1^2)-(x+1)/x(x-2)

=x+1/(x+1)(x-1)

=1/(x-1)