\(\Delta=\left[-2\left(m-1\right)\right]^2-4\left(m-2\right)\left(m+1\right)\)
\(=4m^2-8m+4-4\left(m^2+m-2m-2\right)\)
\(=4m^2-8m+4-4m^2+4m+8\)
\(=-4m+12\)
Để pt có 2 nghiệm thì \(\Delta>0\)
\(\Leftrightarrow-4m+12>0\)
\(\Leftrightarrow m< 3\)
Theo hệ thức Vi-ét, ta có:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2m-2}{m+1}\\x_1x_2=\dfrac{m-2}{m+1}\end{matrix}\right.\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{2m-2}{m+1}:\dfrac{m-2}{m+1}=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{2m-2}{m-2}=\dfrac{7}{4}\)
\(\Leftrightarrow8m-8=7m-14\)
\(\Leftrightarrow m=-6\left(tm\right)\)
Vậy \(m=-6\)