Question

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

giải phương trình

Answer 1

ĐKXĐ: \(x\ne\pm3\)

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

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

\(\Rightarrow x-3+2\left(x+3\right)=x^2+3x\)

\(\Leftrightarrow3x+3=x^2+3x\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x=\pm\sqrt{3}\) (thỏa mãn)