a: \(\text{Δ}=\left(2m+2\right)^2-4\left(m^2+2\right)\)
\(=4m^2+8m+4-4m^2-8=8m-4\)
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: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)=2m+2\\x_1x_2=\dfrac{c}{a}=m^2+2\end{matrix}\right.\)
\(x_1^2+x_2^2=10\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2=10\)
=>\(\left(2m+2\right)^2-2\left(m^2+2\right)=10\)
=>\(4m^2+8m+4-2m^2-4-10=0\)
=>\(2m^2+8m-10=0\)
=>(m+5)(m-1)=0
=>\(\left[{}\begin{matrix}m=-5\left(loại\right)\\m=1\left(nhận\right)\end{matrix}\right.\)
b: \(\sqrt{x_1+x_2}+\sqrt{x_1x_2+1}=4\)
=>\(\sqrt{2m+2}+\sqrt{m^2+2+1}=4\)
=>\(\sqrt{m^2+3}+\sqrt{2m+2}-4=0\)
=>\(\sqrt{m^2+3}-2+\sqrt{2m+2}-2=0\)
=>\(\dfrac{m^2+3-4}{\sqrt{m^2+3}+2}+\dfrac{2m+2-4}{\sqrt{2m+2}+2}=0\)
=>\(\left(m-1\right)\left(\dfrac{m+1}{\sqrt{m^2+3}+2}+\dfrac{2}{\sqrt{2m+2}+2}\right)=0\)
=>m-1=0
=>m=1(nhận)
