\(6n^2-5n-1\)\(\text{ là hợp số }\left(n\in N\right)\)

\(\text{Chứng minh :}\)\(6n^2-5n-1\) \(\text{là hợp số}\)

\(\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\)

\(\text{CMR: }x^{2019}+y^{2019}+z^{2019}=\left(x+y+z\right)^{2019}\)

Tìm x, y nguyên thỏa mãn:

\(x^2+2y^2+2xy-2x+4y=10\)

Tìm GTLN:

\(\frac{\sqrt{x-2019}}{x+2}+\frac{\sqrt{x-2020}}{x}\)