\(\dfrac{m}{x+1}+\dfrac{n}{x-2}=\dfrac{3.2x-19}{x^2-x-2}\) với \(x\ne-1;2\)
\(\Leftrightarrow\dfrac{m\left(x-2\right)+n\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{6x-19}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow m\left(x-2\right)+n\left(x+1\right)=6x-19\)
\(\Leftrightarrow\left(m+n\right)x+\left(n-2m\right)=6x-19\)
\(\Leftrightarrow\left\{{}\begin{matrix}m+n=6\left(1\right)\\n-2m=-19\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow n=6-m\) Thay vào (2) ta có :
\(6-m-2m=-19\)
\(\Leftrightarrow\) \(m=\dfrac{25}{3}\) \(\Rightarrow\) n=\(\dfrac{-7}{3}\)
Vậy m.n=\(\dfrac{25}{3}.\dfrac{-7}{3}=\dfrac{-175}{9}\)
