1) \(x^2+\left(m+1\right)x+m=2\) ( 1 )
\(pt\left(1\right)\Leftrightarrow x^2+\left(m+1\right)+m-2=0\)
\(\Delta=b^2-4ac\)
\(\Delta=\left(m-1\right)^2+8\ge8\) \(\forall m\in R\)
\(\Rightarrow\) đpcm
2)
Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=-m-1\\x_1x_2=m-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_1+x_2\right)^2=\left(-m-1\right)^2\\2x_1x_2=2m-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+2x_1x_2+x^2_2=m^2+2m+1\\2x_1x_2=2m-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2+2m-4=m^2+2m+1\\2x_1x_2=2m-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2=\left(m^2+2m+1\right)-\left(2m-4\right)\\2x_2x_1=2m-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2=m^2+5\\x_1x_2=m-2\end{matrix}\right.\)
Theo yêu cầu đề bài \(\dfrac{2x_1-1}{x_2}+\dfrac{2x_2-1}{x_1}=x_1x_2+\dfrac{55}{x_1x_2}\)
\(\Leftrightarrow\dfrac{2\left(x^2_1+x^2_2\right)-\left(x_1+x_2\right)}{x_1x_2}=x_1x_2+\dfrac{55}{x_1x_2}\)
\(\Leftrightarrow\dfrac{2\left(m^2+5\right)-\left(-m-1\right)}{m-2}=m-2+\dfrac{55}{m-2}\)
\(\Leftrightarrow\dfrac{2m^2+m+11}{m-2}=\dfrac{\left(m-2\right)^2+55}{m-2}\)
\(\Leftrightarrow2m^2+m+11=\left(m-2\right)^2+55\) ( điều kiện \(me2\) )
\(\Leftrightarrow m^2+5m-48=0\)
\(\Delta=b^2-4ac\)
\(\Rightarrow\Delta=217\)
\(\Rightarrow m_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-5\pm\sqrt{217}}{2}\)
