\(b,\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{x^2-9}\left(đkxđ:x\ne\pm3,x\ne-\dfrac{7}{2}\right)\)
\(\Leftrightarrow\dfrac{13\left(x+3\right)+x^2-9-6\left(2x+7\right)}{\left(x^2-9\right)\left(2x+7\right)}=0\)
\(\Leftrightarrow13x+39+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=3\left(l\right)\\x_2=-4\left(n\right)\end{matrix}\right.\)
Vậy \(S=\left\{-4\right\}\)
\(a,\dfrac{2x-5}{x+3}+\dfrac{2x}{x-1}=\dfrac{4}{x^2+2x-3}\left(đkxđ:x\ne-3;1\right)\)
\(\Rightarrow\dfrac{\left(2x-5\right)\left(x-1\right)+2x\left(x+3\right)-4}{\left(x+3\right)\left(x-1\right)}=0\)
\(\Rightarrow2x^2-2x-5x+5+2x^2+6x-4=0\)
\(\Rightarrow4x^2-x+1=0\)
\(\Rightarrow\) Pt vô nghiệm
Vậy \(S=\varnothing\)
