Question

\(\dfrac{2x+3}{x+1}-\dfrac{6}{X}=2\)

Answer 1

\(\frac{2x+3}{x+1}-\frac{6}{x}=2\) 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 = \(\frac{-6}{5}\)

Answer 2

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}