Bài 7:
\(x^2-\left(m^2+1\right)x-3=0\)
\(\Delta=\left[-\left(m^2+1\right)\right]^2-4\cdot1\cdot\left(-3\right)=\left(m^2+1\right)^2+12>0\forall m\)
⇒ Pt luôn có 2 nghiệm phân biệt
Theo vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-\left[-\left(m^2+1\right)\right]}{1}=m^2+1\\x_1x_2=\dfrac{-3}{1}=-3\end{matrix}\right.\)
Ta có: \(\left|x_1\right|+\left|x_2\right|=4\)
\(\Rightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=4^2\) (vì cả hai vế luôn lớn hơn hoặc bằng 0)
\(\Leftrightarrow\left(\left|x_1\right|\right)^2+\left(\left|x_2\right|\right)^2+2\left|x_1\right|\cdot\left|x_2\right|=16\)
\(\Leftrightarrow\left(x_1^2+x_2^2\right)+2\left|x_1x_2\right|=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=16\)
\(\Rightarrow\left(m^2+1\right)^2-2\cdot\left(-3\right)+2\cdot\left|-3\right|=16\)
\(\Leftrightarrow\left(m^2+1\right)^2+12=16\)
\(\Leftrightarrow\left(m^2+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}m^2+1=2\\m^2+1=-2\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow m^2=1\Leftrightarrow m=\pm1\)
Vậy: ...
