Question

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

Answer 1

\(\dfrac{x+3}{x-3}+\dfrac{x-3}{x+3}=\dfrac{x^2+2x}{x^2-9}+1\left(ĐLXĐ:x\ne3;x\ne-3\right)\)

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

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

\(\Leftrightarrow x^2+3x+3x+9+x^2-3x-3x+9=x^2+2x\)

\(\Leftrightarrow x^2+x^2-x^2+3x+3x-3x-3x-2x+9+9=0\)

\(\Leftrightarrow-2x-18=0\)

\(\Leftrightarrow-2x=18\)

\(\Leftrightarrow x=-9\left(nhận\right)\)

Vậy \(S=\left\{-9\right\}\)