Chuyển vế :

\(x_1^2=2\left(m+1\right)x_1-m^2+1\)

thay vào Phuogw trình tìm m thôi

1. Với m=5

\(\Rightarrow x^2-\left(2.5+1\right).x+5^2-1=0\\ \Rightarrow x^2-11.x=-24\\ \)

\(\Rightarrow x^2-\frac{11}{2}.2.x+\left(\frac{11}{2}\right)^2=-24-\left(\frac{11}{2}\right)^2=\frac{-217}{4}\\ \Rightarrow\left(x+\frac{11}{2}\right)^2=-\frac{217}{4}\)

nên x thuộc rỗng

\(x^2m-2\left(m-1\right)x+m+1=0\)

\(\Delta=b^2-4ac\)

\(\Rightarrow\Delta=4m+4\)

Để phương trình có 2 nghiệm phân biệt 

\(\Rightarrow\Delta>0\Leftrightarrow m>-1\)

Theo định lý Viet 

\(\Rightarrow\hept{\begin{cases}x_1+x_2=\frac{-b}{a}\\x_1x_2=\frac{c}{a}\end{cases}}\) 

 \(\Leftrightarrow\hept{\begin{cases}x_1+x_2=\frac{2\left(m-1\right)}{m}\\x_1.x_2=\frac{m+1}{m}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(x_1+x_2\right)^2=\left[\frac{2\left(m-1\right)}{m}\right]^2\\2x_1x_2=\frac{2\left(m+1\right)}{m}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x_1^2+x_2^2+2x_1x_2=\frac{4\left(m-1\right)^2}{m^2}\left(1\right)\\2x_1x_2=\frac{2\left(m+1\right)}{m}\end{cases}}\)

Xét phương trình ( 1 )

\(pt\left(1\right)\Leftrightarrow16+\frac{2\left(m+1\right)}{m}=\frac{4\left(m-1\right)^2}{m^2}\)

\(\Leftrightarrow\frac{16m+2\left(m+1\right)}{m}=\frac{4\left(m-1\right)^2}{m^2}\)

\(\Leftrightarrow\frac{18m+2}{m}=\frac{4\left(m^2-2m+1\right)}{m^2}\)

\(\Leftrightarrow m^2\left(18m+2\right)=4m\left(m^2-2m+1\right)\)với m khác 0

\(\Leftrightarrow m\left(18m+2\right)=4\left(m^2-2m+1\right)\)

\(\Leftrightarrow18m^2+2m=4m^2-8m+4\)

\(\Leftrightarrow14m^2+10m-4=0\)

\(\Delta=b^2-4ac\)

\(\Rightarrow\Delta=324\)

\(\Rightarrow\hept{\begin{cases}m_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-10+\sqrt{324}}{28}\\m_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-10-\sqrt{324}}{28}\end{cases}}\)

Do  \(m>-1\)

\(\Rightarrow m=\frac{-10+\sqrt{324}}{28}\)