\(\dfrac{a+b}{3a-b}+\dfrac{1}{a+b}.\left(\dfrac{a^2-b^2}{3a-b}\right)\)
ĐKXĐ: \(a;b\ne0\)
\(\dfrac{a+b}{3a-b}+\dfrac{1}{a+b}.\dfrac{\left(a-b\right)\left(a+b\right)}{3a-b}\)
\(=\dfrac{a+b}{3a-b}+\dfrac{a-b}{3a-b}\)
\(=\dfrac{a+b-a+b}{3a-b}=\dfrac{2b}{3a-b}\)
Học tốt nha<3