ĐKXĐ: x∉{-5;5}
Ta có: \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{7}{6x+30}\)
\(\Leftrightarrow\frac{3}{4\left(x-5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}-\frac{7}{6\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}-\frac{90}{12\left(x-5\right)\left(x+5\right)}-\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\)
\(\Leftrightarrow9\left(x+5\right)-90-14\left(x-5\right)=0\)
\(\Leftrightarrow9x+45-90-14x+70=0\)
\(\Leftrightarrow-5x+25=0\)
\(\Leftrightarrow-5x=-25\)
hay x=5(ktm)
Vậy: x∈∅
