a=1; b=m; c=-3
Vì ac<0 nên phương trình luôn có hai nghiệm phân biệt
Theo đề, ta có: \(\left|x_1\right|+\left|x_2\right|=4\)
\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=16\)
\(\Leftrightarrow m^2-2\cdot\left(-3\right)+2\cdot3=16\)
\(\Leftrightarrow m\in\left\{2;-2\right\}\)
\(ac=-3< 0\Rightarrow\) phương trình luôn có 2 nghiệm pb trái dấu
Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-3\end{matrix}\right.\)
\(\left|x_1\right|+\left|x_2\right|=4\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=16\)
\(\Leftrightarrow m^2-2.\left(-3\right)+2\left|-3\right|=16\)
\(\Leftrightarrow m^2+12=16\Rightarrow m=\pm2\)
\(\Delta=m^2+12>0\forall m\\ \Rightarrow Pt.luôn.có.2.n_o.phân.biệt\\ Theo.Viét:\\ \left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-3\end{matrix}\right.\left(1\right)\\ Theo.đb:\\ \left|x_1\right|+\left|x_2\right|=4\\ \Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=4^2\\ \Leftrightarrow\left|x_1\right|^2+\left|x_2\right|^2+2\left|x_1x_2\right|\\ \Leftrightarrow x_1+x_2+2\left|x_1x_2\right|=16\)
\(\Leftrightarrow\left(x_1\right)^2+2x_1x_2+x_2-2x_1x_2+2\left|x_1x_2\right|=16\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=16_{\left(2\right)}\\ Thay\left(1\right)vào\left(2\right):\left(-m\right)^2-2\left(-3\right)+2\left|-3\right|=16\\ \Leftrightarrow m^2+6+6=16\\ \Leftrightarrow m^2=4\Rightarrow m=\pm2\)
