Bài 5 trang 56 SGK Giải tích 12

Chứng minh rằng:

a) \(\left ( \dfrac{1}{3} \right )^{2\sqrt{5}}\) < \(\left ( \dfrac{1}{3} \right )^{3\sqrt{2}}\);       

b) \(7^{\sqrt[6]{3}}\) > \(7^{\sqrt[3]{6}}\)

a) \({\left( {\dfrac{1}{3}} \right)^{2\sqrt 5 }} < {\left( {\dfrac{1}{3}} \right)^{3\sqrt 2 }}.\)
Ta có: \(2\sqrt 5 = \sqrt {{2^2}.5} = \sqrt {20} ;\;\;3\sqrt 2 = \sqrt {{3^2}.2} = \sqrt {18} .\)
Vì \(20 > 18 \Rightarrow \sqrt {20} > \sqrt {18} \Leftrightarrow 2\sqrt 5 > 3\sqrt 2 .\)
Lại có: \(0 < \dfrac{1}{3} < 1 \Rightarrow {\left( {\dfrac{1}{3}} \right)^{2\sqrt 5 }} < {\left( {\dfrac{1}{3}} \right)^{3\sqrt 2 }}\) (đpcm)
b) \({7^{6\sqrt 3 }} > {7^{3\sqrt 6 }}.\)
Ta có: \(6\sqrt 3 = \sqrt {{6^2}.3} = \sqrt {108} ;\;\;3\sqrt 6 = \sqrt {{3^2}.6} = \sqrt {54} .\)
Vì \(108 > 54 \Rightarrow 6\sqrt 3 > 3\sqrt 6 .\)
Mà \(7 > 1 \Rightarrow {7^{6\sqrt 3 }} > {7^{3\sqrt 6 }}\) (đpcm)