Question

计算：（1）(ab-b^2)÷a^2+b^2分之a^2-b^2·b分之1 （2）x^2+3x+2分之x+1 + x^2+

Answer 1

1）(ab-b^2)÷[(a^2-b^2)/(a^2+b^2)]*(1/b)

=b(a-b)*[(a^2+b^2)/(a+b)(a-b)]*1/b

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

2)x^2+3x+2分之x+1 + x^2+7x+12分之x+3

=(x+1)/[(x+1)(x+2)]+(x+3)/[(x+3)(x+4)]

=1/(x+2)+1/(x+4)

=[(x+4)+(x+2)]/[(x+2)(x+4)]

=(2x+6)/(x+2)(x+4)