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Câu hỏi: Cho $P={{\left( 5-2\sqrt{6} \right)}^{2018}}{{\left( 5+2\sqrt{6} \right)}^{2019}}$. Khẳng định nào sau đây đúng?
A. $P\in \left( 2;7 \right)$.
B. $P\in \left( 6;9 \right)$.
C. $P\in \left( 0;3 \right)$.
D. $P\in \left( 8;10 \right)$.
A. $P\in \left( 2;7 \right)$.
B. $P\in \left( 6;9 \right)$.
C. $P\in \left( 0;3 \right)$.
D. $P\in \left( 8;10 \right)$.
Ta có $P={{\left( 5-2\sqrt{6} \right)}^{2018}}{{\left( 5+2\sqrt{6} \right)}^{2019}}={{\left( 5-2\sqrt{6} \right)}^{2018}}{{\left( 5+2\sqrt{6} \right)}^{2018}}\left( 5+2\sqrt{6} \right)$
$=\left[ {{\left( 5-2\sqrt{6} \right)}^{2018}}{{\left( 5+2\sqrt{6} \right)}^{2018}} \right]\left( 5+2\sqrt{6} \right)$
$={{\left[ \left( 5-2\sqrt{6} \right)\left( 5+2\sqrt{6} \right) \right]}^{2018}}\left( 5+2\sqrt{6} \right)={{\left( 1 \right)}^{2018}}\left( 5+2\sqrt{6} \right)=5+2\sqrt{6}$
Vậy $P\in \left( 8;10 \right)$.
$=\left[ {{\left( 5-2\sqrt{6} \right)}^{2018}}{{\left( 5+2\sqrt{6} \right)}^{2018}} \right]\left( 5+2\sqrt{6} \right)$
$={{\left[ \left( 5-2\sqrt{6} \right)\left( 5+2\sqrt{6} \right) \right]}^{2018}}\left( 5+2\sqrt{6} \right)={{\left( 1 \right)}^{2018}}\left( 5+2\sqrt{6} \right)=5+2\sqrt{6}$
Vậy $P\in \left( 8;10 \right)$.
Đáp án D.