\(\text{Δ}=\left(2m+2\right)^2-4\left(m^2+2\right)\)
\(=4m^2+8m+4-4m^2-8=8m-4\)
Để phương trinh có 2 nghiệm phân biệt thì Δ>0
=>8m-4>0
=>8m>4
=>\(m>\dfrac{1}{2}\)
Theo Vi-et, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)=2m+2\\x_1x_2=\dfrac{c}{a}=m^2+2\end{matrix}\right.\)
\(\sqrt{x_1+x_2}+\sqrt{x_1x_2+1}=4\)
=>\(\sqrt{2m+2}+\sqrt{m^2+2+1}=4\)
=>\(\sqrt{m^2+3}+\sqrt{2m+2}-4=0\)
=>\(\sqrt{m^2+3}-2+\sqrt{2m+2}-2=0\)
=>\(\dfrac{m^2+3-4}{\sqrt{m^2+3}+2}+\dfrac{2m+2-4}{\sqrt{2m+2}+2}=0\)
=>\(\dfrac{\left(m-1\right)\left(m+1\right)}{\sqrt{m^2+3}+2}+\dfrac{2m-2}{\sqrt{2m+2}+2}=0\)
=>\(\left(m-1\right)\left(\dfrac{m+1}{\sqrt{m^2+3}+2}+\dfrac{2}{\sqrt{2m+2}+2}\right)=0\)
=>m-1=0
=>m=1(nhận)
