ĐKXĐ : \(\left\{{}\begin{matrix}x+5\ne0\\x-5\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-5\\x\ne5\end{matrix}\right.\)
Ta có : \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}\)
=> \(\frac{3}{4\left(x-5\right)}-\frac{15}{2\left(x^2-25\right)}=-\frac{7}{6\left(x+5\right)}\)
=> \(\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}-\frac{90}{12\left(x-5\right)\left(x+5\right)}=-\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}\)
=> \(9\left(x+5\right)-90=-14\left(x-5\right)\)
=> \(9x+45-90+14x-70=0\)
=> \(23x=115\)
=> \(x=5\) ( không thỏa mãn )
Vậy phương trình trên vô nghiệm .
