\(\Leftrightarrow\frac{\left(m+2\right)\left(m+1\right)+\left(m-2\right)}{\left(m-2\right)\left(m+1\right)}=3\)
\(\Leftrightarrow m^2+2m+m+2+m-2=3\left(m^2-2m+m-2\right)\)
\(\Leftrightarrow m^2+4m=3m^2-3m-6\)
\(\Leftrightarrow2m^2-7m-6=0\Rightarrow\left[{}\begin{matrix}m=\frac{7+\sqrt{97}}{4}\\m=\frac{7-\sqrt{97}}{4}\end{matrix}\right.\)
Giải:
Ta có: \(\frac{m+2}{m-2}+\frac{1}{m+1}=3\) (ĐKXĐ: m ≠ 2; m ≠ -1)
\(\Leftrightarrow\frac{m^2+3m+2}{\left(m-2\right)\left(m+1\right)}+\frac{m-2}{\left(m-2\right)\left(m+1\right)}=3\)
\(\Leftrightarrow\frac{m^2+4m}{m^2-m-2}=3\)
\(\Leftrightarrow m^2+4m=3m^2-3m-6\)
\(\Leftrightarrow2m^2-7m-6=0\)
tịt -.-