\(x+8⋮x^2+1\)
Mà \(x^2+1⋮x^2+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+8x⋮x^2+1\\x^2+1⋮x^2+1\end{matrix}\right.\)
\(\Leftrightarrow8x-1⋮x^2+1\)
Mà \(x+8⋮x^2+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-1⋮x^2+1\\8x+64⋮x^2+1\end{matrix}\right.\)
\(\Leftrightarrow65⋮x^2+1\)
\(\Leftrightarrow x^2+1\inƯ\left(65\right)\)
Tự xét nhé!
ta có (x+8)\(⋮\)(x\(^2\)+1)
=>(x+8)(x-8)\(⋮\)(x\(^2\)+1)
=>(x\(^2\)-64)\(⋮\)(x\(^2\)+1)
=>(x\(^2\)+1)+65\(⋮\)(x\(^2\)+1)
=>65\(⋮\)(x\(^2\)+1)
=>(x\(^2\)+1)\(\in\)U(65)
=>(x\(^2\)+1)\(\in\){\(\pm1;\pm65\)}
=>x\(^2\)\(\in\){0;64}
=>x\(\in\){0;\(\pm\)8}
