c)
ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{2x^2-5}{x^3-1}-\dfrac{4}{x^2+x+1}=\dfrac{1}{1-x}\)
\(\Leftrightarrow\dfrac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
Suy ra: \(2x^2-5-4x+4+x^2+x+1=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
d)
ĐKXĐ: \(x\notin\left\{3;-3;-\dfrac{7}{2}\right\}\)
Ta có: \(\dfrac{6}{x^2-9}=\dfrac{1}{2x+7}-\dfrac{13}{\left(2x+7\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=\dfrac{x^2-9}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{13\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(12x+42=x^2-9+13x+39\)
\(\Leftrightarrow x^2+13x+30-12x-42=0\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-4}
