Đặt a=x-7\(\Leftrightarrow a-1=x-8\)
Vậy \(\left(x-7\right)^4+\left(x-8\right)^4=\left(15-2x\right)^4\Leftrightarrow a^4+\left(a-1\right)^4=\left(1-2a\right)^4\Leftrightarrow a^4+a^4-2a^3+a^2-2a^3+4a^2-2a+a^2-2a+1=16a^4-16a^3+4a^2-16a^3+16a^2-4a+4a^2-4a+1\Leftrightarrow2a^4-4a^3+4a^2-4a+1=16a^4-32a^3+24a^2-8a+1\Leftrightarrow14a^4-28a^3+18a^2-4a=0\Leftrightarrow7a^4-14a^3+9a^2-2a=0\Leftrightarrow a\left(7a^3-14a^2+9a-2\right)=0\Leftrightarrow a\left(a-1\right)\left(7a^2-7a+2\right)=0\left(1\right)\)
Vì \(7a^2-7a+2>0\)
Vậy (1)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=0\\a=1\end{matrix}\right.\)
_ a=0\(\Leftrightarrow x-7=0\Leftrightarrow x=7\)
_ a=1\(\Leftrightarrow x-7=1\Leftrightarrow x=8\)
Vậy S={7;8}
