Question

[Duong Thi Nhuong]

Rút gọn :

a) \(\left(a-\frac{a^2+b^2}{a-b}\right).\left(\frac{1}{b}+\frac{2}{a+b}\right)\)

b) \(\left(\frac{3a+1}{a^2-3a}+\frac{3a-1}{a^2+3a}\right).\frac{a^2-9}{a^2+1}\)

Answer 1

[Nguyen Thi Trinh]

a/ đk: a\(\ne b\), b\(\ne0,a\ne-b\)

= \(\frac{a\left(a-b\right)-a^2-b^2}{a-b}.\frac{a+b+2b}{b\left(a+b\right)}\)

= \(\frac{a^2-ab-a^2-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)

= \(\frac{-ab-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)

= \(\frac{-b\left(a+b\right)\left(a+3b\right)}{b\left(a+b\right)\left(a-b\right)}\)

= \(\frac{-a-3b}{a-b}\)

b/ đk: a\(\ne0,a\ne\pm3\)

= \(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right].\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{\left(3a+1\right)\left(a+3\right)+\left(3a-1\right)\left(a-3\right)}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)

= \(\frac{6\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}{a\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}\)

= \(\frac{6}{a}\)