Ta có :
\(\left(x-1\right)^4+\left(5-x\right)^4=1^4+3^4\)
\(\Rightarrow\hept{\begin{cases}x-1=2\\5-x=3\end{cases}}\)hoặc\(\Rightarrow\hept{\begin{cases}x-1=3\\5-x=1\end{cases}}\)
\(\Rightarrow x=2\)hoặc\(\Rightarrow x=4\)
Vậy, \(\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
\(\left(x-1\right)^4+\left(5-x\right)^4=82\)
\(\Leftrightarrow\left(x-1\right)^4+\left(x-5\right)^4=82\)
Đặt \(x-3=y\Rightarrow x=y+3\)
Thay \(x=y+3\)vào phương trình. Ta có:
\(\left(y+2\right)^4+\left(y-2\right)^4=82\)
\(\Leftrightarrow y^4+8y^3+24y^2+32y+16+y^4-8y^3+24y^2-32y+16=82\)
\(\Leftrightarrow2y^4+48y^2+32=82\)
\(\Leftrightarrow2y^4+48y^2+32-82=0\)
\(\Leftrightarrow2y^4+48y^2-50=0\)
\(\Leftrightarrow2\left(y^2-1\right)\left(y^2+25\right)=0\)
\(\Leftrightarrow2\left(y-1\right)\left(y+1\right)\left(y^2+25\right)=0\)
\(\orbr{\begin{cases}\orbr{\begin{cases}y-1=0\\y+1=0\end{cases}}\\y^2+25=0\left(y^2+25\ge25>0\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=1\\y=-1\end{cases}}\)\(\Rightarrow y=1\)hoặc \(y=-1\)
Nếu \(y=1\Rightarrow x=4\)
Nếu\(y=-1\Rightarrow x=2\)
Vậy x=4 hoặc x=2
