\(\Delta=\left(m+1\right)^2-4m=m^2+2m+1-4m=m^2-2m+1=\left(m-1\right)^2\ge0\)
=>Phương trình luôn có hai nghiệm
Theo Vi-et, ta có:
\(x_1+x_2=-\frac{b}{a}=m+1;x_1\cdot x_2=\frac{c}{a}=m\)
Để \(\sqrt{x_1};\sqrt{x_2}\) tồn tại thì \(\begin{cases}x_1+x_2>0\\ x_1x_2>0\end{cases}\Rightarrow\begin{cases}m+1>0\\ m>0\end{cases}\)
=>m>0
Ta có: \(\sqrt{x_1}+\sqrt{x_2}=4\)
=>\(x_1+x_2+2\sqrt{x_1x_2}=16\)
=>\(m+1+2\sqrt{m}=16\)
=>\(m+2\sqrt{m}-15=0\)
=>\(\left(\sqrt{m}+5\right)\left(\sqrt{m}-3\right)=0\)
mà \(\sqrt{m}+5\ge5>0\forall m>0\)
nên \(\sqrt{m}-3=0\)
=>m=9(nhận)
=>\(x_1+x_2=m+1=9+1=10;x_1x_2=\frac{c}{a}=m=9\)
\(x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=10^3-3\cdot9\cdot10=730\)
