\(\frac{2x+3}{x+1}-\frac{6}{x}=2\) ĐKXĐ x khác 0 và x khác -1
<=> \(\frac{x.\left(2x+3\right)}{x.\left(x+1\right)}-\frac{6.\left(x+1\right)}{x.\left(x+1\right)}=\frac{2x.\left(x+1\right)}{x.\left(x+1\right)}\)
=> \(2x^2+3x-6x-6=2x^2+2x\)
<=> \(2x^2-2x^2+3x-6x-2x=6\)
<=> -5x = 6
<=> x = \(\frac{-6}{5}\)
ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\frac{2x+3}{x+1}-\frac{6}{x}=2\)
\(\Leftrightarrow\frac{x\left(2x+3\right)}{x\left(x+1\right)}-\frac{6\left(x+1\right)}{x\left(x+1\right)}=\frac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(2x^2+6x-6x-6=2x^2+2x\)
\(\Leftrightarrow2x^2-6-2x^2-2x=0\)
\(\Leftrightarrow-6-2x=0\)
\(\Leftrightarrow6+2x=0\)
\(\Leftrightarrow2x=-6\)
hay x=-3(tm)
Vậy: S={-3}