Theo định lý Vi ét phương trình cho ta có : \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m\end{matrix}\right.\)
Theo đề bài ta có \(\sqrt{x_1}+\sqrt{x_2}=3\left(x_1;x_2\ge0\right)\left(1\right)\)
Để phương trình cho có \(2\) nghiệm \(x_1;x_2>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(m+1\right)^2-4m>0\\S=m+1>0\\P=m>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)^2>0\\m>-1\\m>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\m\ne1\end{matrix}\right.\left(2\right)\)
\(\left(1\right)\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2}=9\)
\(\Leftrightarrow m+1+2\sqrt{m}=9\)
\(\Leftrightarrow2\sqrt{m}=8-m\)
\(\Leftrightarrow4m=64-16m+m^2\left(m\le8\right)\)
\(\Leftrightarrow m^2-20m+64=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=4\\m=16\end{matrix}\right.\) so với \(\left(2\right)\&m\le8\Rightarrow m=4\Rightarrow\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=4\end{matrix}\right.\)
\(x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=5^3-3.4.5=65\)
