Question

giải PT:

a/\(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)

Answer 1

a/

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

\(\Leftrightarrow\dfrac{\left(x^2-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2.\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-\left(7x^2-3x\right)}{\left(x-3\right)\left(x+3\right)}\) \(\Leftrightarrow x^3-4x^2+3x-x^3-3x^2=7x^2+3x\)

\(\Leftrightarrow x^3-4x^2+3x-x^3-3x^2-7x^2-3x=0\)

\(\Leftrightarrow-14x^2=0\)

\(\Leftrightarrow x=0\)