Question

Thực hiện phép cộng:  

\(\dfrac{x-1}{3x+3}+\dfrac{3x}{x^2-1}\) 

Answer 1

\(=\dfrac{x^2-2x+1+9x}{3\left(x+1\right)\left(x-1\right)}=\dfrac{x^2+7x+1}{3\left(x+1\right)\left(x-1\right)}\)

Answer 2

`[x-1]/[3x+3]+[3x]/[x^2-1]`           `ĐK: x \ne +-1`

`=[(x-1)^2+9x]/[3(x+1)(x-1)]`

`=[x^2-2x+1+9x]/[3(x+1)(x-1)]`

`=[x^2+7x+1]/[3(x+1)(x-1)]`