a: ĐKXĐ: x<>1; x<>-1
\(\Leftrightarrow\left(x-m\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow x^2+x-mx-m+x^2-3x+2-2x^2+2=0\)
\(\Leftrightarrow-2x-mx-m+4=0\)
=>x(-m-2)=m-4
Để PT VN thì -m-2=0
=>m=-2
b: ĐKXĐ: x<>1; x<>m
\(\Leftrightarrow\left(x+1\right)\left(x-m\right)=\left(x+2\right)\left(x-1\right)\)
=>x^2-xm+x-m=x^2+x-2
=>-xm+x-m=x+2
=>-xm-m=2
=>-xm=m+2
=>xm=-m-2
Để PT có nghiệm duy nhất thì m<>0