\(\Delta'=\left(m+1\right)^2-\left(m^2+3\right)=2m-2\ge0\Rightarrow m\ge1\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+3\end{matrix}\right.\)
\(\left|x_1\right|+\left|x_2\right|=10\)
\(\Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=100\)
\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=100\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=100\)
\(\Leftrightarrow4\left(m+1\right)^2-2\left(m^2+3\right)+2\left|m^2+3\right|=100\)
\(\Leftrightarrow4\left(m+1\right)^2-2\left(m^2+3\right)+2\left(m^2+3\right)=100\) (do \(m^2+3>0;\forall m\))
\(\Leftrightarrow4\left(m+1\right)^2=100\)
\(\Leftrightarrow\left(m+1\right)^2=25\Rightarrow\left[{}\begin{matrix}m=4\\m=-6< 1\left(loại\right)\end{matrix}\right.\)
//Hoặc ta có thể biện luận:
Do \(x_1x_2=m^2+3>0;\forall m\) nên \(x_1;x_2\) cùng dấu
\(\Rightarrow\left|x_1\right|+\left|x_2\right|=\left|x_1+x_2\right|\)
Nên: \(\left|x_1\right|+\left|x_2\right|=10\Leftrightarrow\left|x_1+x_2\right|=10\)
\(\Leftrightarrow\left|2\left(m+1\right)\right|=10\)
\(\Leftrightarrow\left|m+1\right|=5\Rightarrow\left[{}\begin{matrix}m=4\\m=-6\left(loại\right)\end{matrix}\right.\)
