Question

\(\dfrac{2x+5}{\left(x+1\right)\left(x+2\right)}=\dfrac{3}{X+1}-\dfrac{1}{x+2}\)

Answer 1

ĐKXĐ: \(x\notin\left\{-1;-2\right\}\)

Ta có: \(\dfrac{2x+5}{\left(x+1\right)\left(x+2\right)}=\dfrac{3}{x+1}-\dfrac{1}{x+2}\)

\(\Leftrightarrow\dfrac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+2\right)}=\dfrac{2x+5}{\left(x+1\right)\left(x+2\right)}\)

Suy ra: \(3x+6-x-1=2x+5\)

\(\Leftrightarrow2x+5-2x-5=0\)

\(\Leftrightarrow0x=0\)(phương trình có vô số nghiệm)

Vậy: S={x|\(x\notin\left\{-1;-2\right\}\)}