\(\dfrac{1+x}{3-x}-\dfrac{1-2x}{3+x}-\dfrac{x\left(1-x\right)}{9-x^2}\)
\(=\dfrac{x^2+4x+3-\left(3-x\right)\left(1-2x\right)}{\left(3-x\right)\left(3+x\right)}-\dfrac{x\left(1-x\right)}{9-x^2}\)
\(=\dfrac{x^2+4x+3-3+7x-2x^2-x+x^2}{9-x^2}\)
\(=\dfrac{10x}{9-x^2}\)
ta có :
\(\dfrac{1+x}{3-x}-\dfrac{1-2x}{3+x}-\dfrac{x\cdot\left(1-x\right)}{9-x^2}\)
= \(\dfrac{\left(1+x\right)\cdot\left(3+x\right)}{\left(3-x\right)\cdot\left(3+x\right)}-\dfrac{\left(1-2x\right)\cdot\left(3-x\right)}{\left(3+x\right)\cdot\left(3-x\right)}-\dfrac{x\cdot\left(1-x\right)}{\left(3-x\right)\cdot\left(3+x\right)}\)
= \(\dfrac{3+x+3x+x^2-3+x+6x-2x^2-x+x^2}{\left(3+x\right)\cdot\left(3-x\right)}\)
= \(\dfrac{10x}{9-x^2}\)
bài lm có j sai sót mong bn thông cảm . 
