ĐKXĐ: \(0\le x\le97\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{97-x}=a\\\sqrt[4]{x}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(pt\Leftrightarrow a+b=5\)
Lại có \(a^4+b^4=97\)
\(\Leftrightarrow\left[\left(a+b\right)^2-2ab\right]^2-2ab=97\)
\(\Leftrightarrow\left(25-2ab\right)^2-2ab=97\)
\(\Leftrightarrow a^2b^2-50ab+264=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ab=44\\ab=6\end{matrix}\right.\)
TH1: \(ab=44\)
Ta có \(\left\{{}\begin{matrix}a+b=5\\ab=44\end{matrix}\right.\Rightarrow\) vô nghiệm
TH2: \(ab=6\)
Ta có \(\left\{{}\begin{matrix}a+b=5\\ab=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt[4]{97-x}=2\\\sqrt[4]{x}=3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt[4]{97-x}=3\\\sqrt[4]{x}=2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=81\left(tm\right)\\x=16\left(tm\right)\end{matrix}\right.\)
Vậy phương trình có hai nghiệm \(x=81;x=16\)
