\(\Delta=\left(m+1\right)^2-4m=m^2+2m+1-4m=m^2-2m+1=\left(m-1\right)^2\ge0\forall m\)
=>Phương trình luôn có hai nghiệm
Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=m+1;x_1x_2=\frac{c}{a}=m\)
Để \(\sqrt{x_1};\sqrt{x_2}\) tồn tại thì \(\begin{cases}x_1+x_2>=0\\ x_1x_2>=0\end{cases}\Rightarrow\begin{cases}m+1>=0\\ m>=0\end{cases}\Rightarrow m\ge0\)
\(\sqrt{x_1}+\sqrt{x_2}=3\)
=>\(x_1+x_2+2\sqrt{x_1x_2}=3^2=9\)
=>\(m+1+2\sqrt{m}=9\)
=>\(m+2\sqrt{m}-8=0\)
=>\(\left(\sqrt{m}+4\right)\left(\sqrt{m}-2\right)=0\)
mà \(\sqrt{m}+4\ge4>0\forall m\)
nên \(\sqrt{m}-2=0\)
=>m=4
=>\(x_1+x_2=4+1=5;x_1x_2=4\)
\(x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=5^3-3\cdot5\cdot4=125-60=65\)
