\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3-3\left(x+\dfrac{1}{x}\right)-3\left(y+\dfrac{1}{y}\right)=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3=15m+5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=u\\y+\dfrac{1}{y}=v\end{matrix}\right.\) với \(\left|u\right|;\left|v\right|\ge2\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=5\\u^3+v^3=15m+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u+v=5\\\left(u+v\right)^3-3uv\left(u+v\right)=15m+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u+v=5\\uv=-m+8\end{matrix}\right.\)
\(\Rightarrow u\left(5-u\right)=-m+8\Leftrightarrow u^2-5u+8=m\)
Ta có: \(v=5-u\) , mà \(\left[{}\begin{matrix}v\ge2\\v\le-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}5-u\ge2\\5-u\le-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u\le3\\u\ge7\end{matrix}\right.\)
Kết hợp \(\left|u\right|\ge2\Rightarrow\left[{}\begin{matrix}u\le-2\\2\le u\le3\\u\ge7\end{matrix}\right.\)
Xét hàm \(f\left(u\right)=u^2-5u+8\) trên \((-\infty;-2]\cup\left[2;3\right]\cup[7;+\infty)\)
Từ đồ thị \(f\left(u\right)\) ta thấy \(y=m\) cắt \(y=f\left(u\right)\) trên miền đã cho khi \(\left[{}\begin{matrix}\dfrac{7}{4}\le m\le2\\m\ge22\\\end{matrix}\right.\)
