\(\left(m+1\right)x^2-2mx+m-5=0\left(1\right)\)
\(TH1:m+1=0\Leftrightarrow m=-1\Rightarrow\left(1\right)\Leftrightarrow x=3\left(ktm\right)\)
\(TH2:m+1\ne0\Leftrightarrow m\ne-1\Rightarrow\left(1\right)\) \(có\) \(2ngo\) \(pb\)
\(\Leftrightarrow\Delta'>0\Leftrightarrow m^2-\left(m-5\right)\left(m+1\right)>0\Leftrightarrow m>-\dfrac{5}{4}\)
\(vi-ét\Rightarrow\left\{{}\begin{matrix}x1+x2=\dfrac{2m}{m+1}\\x1x2=\dfrac{m-5}{m+1}\end{matrix}\right.\)
\(x1+2x2-2x1x2=4\)
\(\Leftrightarrow\dfrac{2m}{m+1}+x2-\dfrac{2\left(m-5\right)}{m+1}=4\)
\(\Leftrightarrow x2+\dfrac{2m-2\left(m-5\right)}{m+1}=4\)
\(\Leftrightarrow x2+\dfrac{2m+10}{m+1}=4\Rightarrow x2=4-\dfrac{2m+10}{m+1}\)
\(=\dfrac{4m+4-2m-10}{m+1}=\dfrac{2m-6}{m+1}\)
\(\Rightarrow x1=\dfrac{2m}{m+1}-\dfrac{2m-6}{m+1}=\dfrac{6}{m+1}\)
\(\Rightarrow\dfrac{6}{m+1}.\dfrac{2m-6}{m+1}=\dfrac{m-5}{m+1}\Leftrightarrow m=8\pm\sqrt{33}\left(tm\right)\)
(kiểm tra lại tính toán giùm )
