Ta có: \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=5\\x^3+y^3+\dfrac{1}{x^3}+\dfrac{1}{y^3}=15m-10\end{matrix}\right.\)\(\left(x,ye0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)+\dfrac{x+y}{xy}=5\\\left(x^3+y^3\right)+\dfrac{x^3+y^3}{x^3y^3}=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(1+\dfrac{1}{xy}\right)=5\\\left(x^3+y^3\right)\left(1+\dfrac{1}{x^3y^3}\right)=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(1+\dfrac{1}{xy}\right)=5\\\left(x+y\right)\left(1+\dfrac{1}{xy}\right)\left(x^2-xy+y^2\right)\left(\dfrac{1}{x^2y^2}-\dfrac{1}{xy}+1\right)=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(1+\dfrac{1}{xy}\right)=5\\5\left(\left(x+y\right)^2-3xy\right)\left(\left(\dfrac{1}{xy}+1\right)^2-\dfrac{2}{xy}\right)=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(1+\dfrac{1}{xy}\right)=5\\\left(x+y\right)^2.\left(1+\dfrac{1}{xy}\right)^2-\dfrac{2\left(x+y\right)^2}{xy}-3xy\left(\dfrac{1}{xy}+1\right)^2+6=3m-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(1+\dfrac{1}{xy}\right)=\dfrac{5}{x+y}\\\dfrac{2\left(x+y\right)^2}{xy}+3xy.\dfrac{25}{\left(x+y\right)^2}=33-3m\end{matrix}\right.\)

Đặt \(t=\dfrac{\left(x+y\right)^2}{xy}\)\(\left(te0\right)\)

Ta có:\(2t+\dfrac{75}{t}=33-3m\)

\(\Leftrightarrow2t^2+\left(3m-33\right)t+75=0\)

Ta có: \(\Delta=\left(3m-33\right)^2-4.2.75=\left(3m-33+10\sqrt{6}\right)\left(3m-33-10\sqrt{6}\right)\)

Để phương trình có nghiệm thì \(\Delta\ge0\)

\(\Rightarrow\left(3m-33+10\sqrt{6}\right)\left(3m-33-10\sqrt{6}\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}m\ge\dfrac{33+10\sqrt{6}}{3}\\m\le\dfrac{33-10\sqrt{6}}{3}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}m\ge\dfrac{33+10\sqrt{6}}{3}\\m\le\dfrac{33-10\sqrt{6}}{3}\end{matrix}\right.\)thì hệ phương trình có nghiệm thực