Đặt \(\hept{\begin{cases}x+5=y\\x-4=z\end{cases}}\)
\(\Leftrightarrow2x+1=y+z\)
=> PT có dạng
\(y^4+z^4=\left(y+z\right)^4\)
\(\Rightarrow y^4+z^4=y^4+4y^3z+6y^2z^2+4yz^3+z^4\)
\(\Leftrightarrow2yz\left(2y^2+3yz+2z^2\right)=0\)
\(\Leftrightarrow2\left(x+5\right)\left(x-4\right)\left(7x^2+7x+22\right)=0\)(1)
Dễ thấy \(7x^2+7x+22=7\left(x+\frac{1}{2}\right)^2+\frac{81}{4}>0\)
Từ \(2\left(x+5\right)\left(x-4\right)\left(7x^2+7x+22\right)=0\)(1)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=4\end{cases}}\)
Vậy .....