Question

[Sách Giáo Khoa]

Hãy so sánh mỗi số sau với 1 :

a) \(2^{-2}\)

b) \(\left(0,013\right)^{-1}\)

c) \(\left(\dfrac{2}{7}\right)^5\)

d) \(\left(\dfrac{1}{2}\right)^{\sqrt{3}}\)

e) \(\left(\dfrac{\pi}{4}\right)^{\sqrt{5}-2}\)

g) \(\left(\dfrac{1}{3}\right)^{\sqrt{8}-3}\)

Answer 1

[Giáo viên Toán]

a) \(2^{-2}=\dfrac{1}{2^2}< 1\)

b) \(\left(0,013\right)^{-1}=\dfrac{1}{0,013}>1\)

c) \(\left(\dfrac{2}{7}\right)^5=\dfrac{2^5}{7^5}< 1\)

d) \(\left(\dfrac{1}{2}\right)^{\sqrt{3}}=\dfrac{1}{2^{\sqrt{3}}}< \dfrac{1}{2^{\sqrt{1}}}=\dfrac{1}{2}< 1\)

e) vì \(0< \dfrac{\pi}{4}< 1\)

Suy ra \(\left(\dfrac{\pi}{4}\right)^{\sqrt{5}-2}=\dfrac{\left(\dfrac{\pi}{4}\right)^{\sqrt{5}}}{\left(\dfrac{\pi}{2}\right)^2}>\dfrac{\left(\dfrac{\pi}{4}\right)^{\sqrt{4}}}{\left(\dfrac{\pi}{4}\right)^2}=1\)

f) Vì \(0< \dfrac{1}{3}< 1\)

Nên \(\left(\dfrac{1}{3}\right)^{\sqrt{8}-3}>\left(\dfrac{1}{3}\right)^{\sqrt{9}-3}=\left(\dfrac{1}{3}\right)^0=1\)