\(x^2+2mx-1=0\)
\(ac=-1.1=-1< 0\Rightarrow\) pt có 2 nghiệm phân biệt
Áp dụng hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-2m\\x_1x_2=-1\end{matrix}\right.\)
Theo đề:\(x_1+2x_2=0\Rightarrow x_1=-2x_2\)
\(\Rightarrow-2x_2^2=-1\Rightarrow x_2^2=\dfrac{1}{2}\Rightarrow x_2=\pm\sqrt{\dfrac{1}{2}}\Rightarrow x_1=\pm\sqrt{2}\)
\(TH_1:\left\{{}\begin{matrix}x_2=\sqrt{\dfrac{1}{2}}\\x_1=-\sqrt{2}\end{matrix}\right.\Rightarrow x_1+x_2=-\dfrac{\sqrt{2}}{2}\Rightarrow m=\dfrac{\sqrt{2}}{4}\)
\(TH_2:\left\{{}\begin{matrix}x_2=-\sqrt{\dfrac{1}{2}}\\x_1=\sqrt{2}\end{matrix}\right.\Rightarrow x_1+x_2=\dfrac{\sqrt{2}}{2}\Rightarrow m=-\dfrac{\sqrt{2}}{4}\)
