Question

[Lê An Bình]

Sắp xếp theo thứ tự giảm dần :

\(\sqrt{2};\left(2^3\right)^{\log_{64}\frac{5}{4}};2^{\frac{\pi}{6}};2^{3\log_92}\)

Answer 1

[Lê Thị Thùy Linh]

Ta có :

\(\sqrt{2}=2^{\frac{1}{2}}\)

\(\left(2^3\right)^{\log_{64}\frac{5}{4}}=2^{3\log_{2^6}\frac{5}{4}}=2^{\frac{1}{2}\log_2\frac{5}{4}}=2^{\log_2\sqrt{\frac{5}{4}}}=\sqrt{\frac{5}{4}}=\left(\frac{5}{4}\right)^{\frac{1}{2}}\)

\(2^{3^{\log_92}}=2^{3^{\frac{1}{2}\log_32}}=2^{3^{\log_3\sqrt{2}}}=2^{\sqrt{2}}\)

Mà : \(\sqrt{2}>\frac{\pi}{6}>\frac{1}{2}\Rightarrow2^{\sqrt{2}}>2^{\frac{\pi}{6}}>2^{\frac{1}{2}}\)

                            \(\Leftrightarrow2^{3^{\log_92}}>2^{\frac{\pi}{6}}>\sqrt{2}\)  (1)

Mặt khác : \(2>\frac{5}{4}\Rightarrow2^{\frac{1}{2}}>\left(\frac{5}{4}\right)^{\frac{1}{2}}\) hay \(\sqrt{2}>\left(2^3\right)^{\log_{64}\frac{5}{4}}\)  (2)

Từ (1) và (2) : \(2^{3^{\log_92}}>2^{\frac{\pi}{6}}>\sqrt{2}>\left(2^3\right)^{\log_{64}\frac{5}{4}}\)

Vậy thứ tự giảm dần là :

\(2^{3^{\log_92}};2^{\frac{\pi}{6}};\sqrt{2};\left(2^3\right)^{\log_{64}\frac{5}{4}}\)