Question

rút gọn P

\(\dfrac{2x}{x+3}-\dfrac{x}{3-x}+\dfrac{6x}{9-x^2}\)

Answer 1

\(\dfrac{2x}{x+3}-\dfrac{x}{3-x}+\dfrac{6x}{9-x^2}=\dfrac{2x\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}-\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(3-x\right)}+\dfrac{6x}{\left(3-x\right)\left(x+3\right)}\)=\(\dfrac{6x-2x^2-x^2-3x+6x}{\left(x+3\right)\left(3-x\right)}=\dfrac{-3x^2+9x}{\left(x+3\right)\left(3-x\right)}\)

=\(\dfrac{3x\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}=\dfrac{3x}{\left(x+3\right)}\)

Answer 2

đk : x\(\ne\) 3 nha mk làm thiếu .