ĐKXĐ: \(x\ne3;x\ne-5\)
\(\frac{2\left|x-3\right|}{x^2+2x-15}=1\Rightarrow2\left|x-3\right|=x^2+2x-15\)
\(\Rightarrow2\left|x-3\right|=\left(x-3\right)\left(x+5\right)\Leftrightarrow\left(x-3\right)\left(x+5\right)-2\left|x-3\right|=0\)
Với \(x>3\) thì \(\Leftrightarrow\left(x-3\right)\left(x+5\right)-2\left(x-3\right)=0\Leftrightarrow\left(x+3\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(KTM\right)\)
Với \(x< 3\) thì: \(\Leftrightarrow\left(x-3\right)\left(x+5\right)+2\left(x-3\right)=0\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\Rightarrow\left[{}\begin{matrix}x=-7\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)
pt có nghiệm duy nhất x=-7
