Question

Tập hợp các số nguyên thỏa mãn là

Answer 1

\(x^{2016}=x^2\)

\(\Rightarrow x^{2016}-x^2=0\)

\(\Rightarrow x^2\left(x^{2014}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x^{2014}-1=0\end{matrix}\right.\)

+) \(x^2=0\Rightarrow x=0\)

+) \(x^{2014}-1=0\)

\(\Rightarrow x^{2014}=1\)

\(\Rightarrow x=\pm1\)

Vậy \(x\in\left\{0;-1;1\right\}\)

Answer 2

x²⁰¹⁶ = x²

=> x²⁰¹⁶ - x² = 0

=> x².x²⁰¹⁴ - x² = 0

=> x².(x²⁰¹⁴ - 1) = 0

=> \(\left[{}\begin{matrix}x^2=0\\x^{2014}-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^{2014}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vậy x = {0;1;-1}