Question

[Duong Thi Nhuong]

Rút gọn  :   \(\left(\frac{1+2a}{4+2a}-\frac{a}{3a-6}+\frac{2a^2}{12-3a^2}\right)\div\frac{13a+6}{24-12a}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

\(=\left(\dfrac{2a+1}{2\left(a+2\right)}-\dfrac{a}{3\left(a-2\right)}-\dfrac{2a^2}{3\left(a-2\right)\left(a+2\right)}\right):\dfrac{13a+6}{24-12a}\)

\(=\dfrac{3\left(2a+1\right)\left(a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}:\dfrac{13a+6}{-12\left(a-2\right)}\)

\(=\dfrac{3\left(2a^2-3a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}\cdot\dfrac{-12\left(a-2\right)}{13a+6}\)

\(=\dfrac{6a^2-9a-6-2a^2-4a-4a^2}{a+2}\cdot\dfrac{-2}{13a+6}\)

\(=\dfrac{-\left(13a+6\right)}{a+2}\cdot\dfrac{-2}{13a+6}=\dfrac{2}{a+2}\)