Đặt \(x^2+2x-3=t\) thì pt trở thành :
\(\frac{24}{t-5}-\frac{15}{t}=2\)
\(\Leftrightarrow\frac{24t-15\left(t-5\right)}{t\left(t-5\right)}=2\)
\(\Leftrightarrow9t+75=2\left(t^2-5t\right)\)
\(\Leftrightarrow9t+75=2t^2-10t\)
\(\Leftrightarrow2t^2-19t-75=0\)
\(\Leftrightarrow2\left(t^2-\frac{19}{2}t-\frac{75}{2}\right)=0\)
\(\Leftrightarrow\left(t-\frac{25}{2}\right)\left(t+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{25}{2}\\t=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=\frac{25}{2}\\x^2+2x-3=-3\end{matrix}\right.\)
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