a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)
\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)
\(\Leftrightarrow3-x-2+7x-7=0\)
\(\Leftrightarrow6x-6=0\)
hay x=1(loại
b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)
\(\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}\)
Suy ra: \(-2-x^2+5x-4=x^2+x-6\)
\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow3-x-2+7x-7=0\)
\(\Rightarrow6x-6=0\)
\(\Rightarrow x=1\)
