Cho $P=\left(5-2\sqrt{6}\right)^{2018}\left(5+2\sqrt{6}\right)^{2019}$. Ta có
$P\in(3;7)$ | |
$P\in(7;9)$ | |
$P\in(9;10)$ | |
$P\in(10;11)$ |
Chọn phương án C.
Ta có $\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=1$.
Suy ra $\left(5-2\sqrt{6}\right)^{2018}\left(5+2\sqrt{6}\right)^{2018}=1$.
Vậy $P=5+2\sqrt{6}\in(9;10)$.