đk : x khác -3
\(\Rightarrow-2x+10=-3\left(x+2\right)\left(x+3\right)\)
\(\Leftrightarrow-2x+10=-3\left(x^2+5x+6\right)\)
\(\Leftrightarrow-2x+10=-3x^2-15x-18\Leftrightarrow3x^2+13x+28=0\)
\(\Delta=169-4.3.28< 0\)
pt vô nghiệm
\(ĐKXĐ:x\ne-3\\ \dfrac{x-5}{3+x}=\dfrac{-3x-6}{-2}\\ \Leftrightarrow\dfrac{x-5}{3+x}=\dfrac{3x+6}{2}\\ \Rightarrow2\left(x-5\right)=\left(3+x\right)\left(3x+6\right)\\ \Leftrightarrow2x-10=9x+3x^2+18+6x\\ \Leftrightarrow3x^2+13x+28=0\\ \Leftrightarrow...\)
Vậy pt vô nghiệm