\(x^2-mx+1=0\left(1\right)\)
Để \(\left(1\right)\) có hai nghiệm phân biệt \(x_1;x_2\) khi
\(\Leftrightarrow\Delta=m^2-4>0\Leftrightarrow m< -2\cup m>2\left(a\right)\)
\(\dfrac{1}{\sqrt{x_1^2+1}+x_1}=\dfrac{\sqrt{x_1^2+1}-x_1}{\left(\sqrt{x_1^2+1}+x_1\right)\left(\sqrt{x_1^2+1}-x_1\right)}=\sqrt{x_1^2+1}-x_1\)
Theo đề bài ta có :
\(\dfrac{1}{\sqrt{x_1^2+1}+x_1}=2\sqrt{2}-x_1-\sqrt{x_2^2+1}\)
\(\Leftrightarrow\sqrt{x_1^2+1}-x_1+x_1+\sqrt{x_2^2+1}=2\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{x_1^2+1}+\sqrt{x_2^2+1}\right)^2=\left(2\sqrt{2}\right)^2\)
\(\Leftrightarrow x_1^2+x_2^2+2\sqrt{\left(x_1^2+1\right)\left(x_2^2+1\right)}+2=8\)
\(\Leftrightarrow x_1^2+x_2^2+2\sqrt{x_1^2x_2^2+x_1^2+x_2^2+1}=6\left(2\right)\)
Theo định lý Vi-ét : \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=1\end{matrix}\right.\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-x_1x_2=m^2-2\)
\(\left(2\right)\Rightarrow m^2-2+2\sqrt{1+m^2-2+1}=6\)
\(\Leftrightarrow2\sqrt{m^2}=8-m^2\)
\(\Leftrightarrow2m=8-m^2\left(\left(a\right)\Rightarrow\left|m\right|>2\right)\)
\(\Leftrightarrow m^2+2m-8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=-4\\m=2\end{matrix}\right.\) so với \(\left(a\right)\Rightarrow m=-4\)
Vậy \(m=-4\) thỏa mãn đề bài
