Question

[títtt]

hãy so sánh mỗi số sau

a) \(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}\) và 1

b) \(\left(\dfrac{1}{5}\right)^{\sqrt{2}}\) và 1

Answer 1

[Nguyễn Việt Lâm]

a.

\(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}=\left(\dfrac{1}{\sqrt{5}}\right)^{-1,2}=\left(5^{-\dfrac{1}{2}}\right)^{-1,2}=5^{\left(-\dfrac{1}{2}\right).\left(-1,2\right)}=5^{0,6}>1\) do \(\left\{{}\begin{matrix}5>1\\0,6>0\end{matrix}\right.\)

b.

\(\left(\dfrac{1}{5}\right)^{\sqrt{2}}=\left(5^{-1}\right)^{\sqrt{2}}=5^{-\sqrt{2}}< 1\) do \(\left\{{}\begin{matrix}5>1\\-\sqrt{2}< 0\end{matrix}\right.\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a: \(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}=\left(\dfrac{1}{\sqrt{5}}\right)^{-\dfrac{6}{5}}=\left(1:\dfrac{1}{\sqrt{5}}\right)^{-\dfrac{5}{6}}=\left(\sqrt{5}\right)^{-\dfrac{5}{6}}\)

\(1=\left(\sqrt{5}\right)^0\)

mà -5/6<0 và \(\sqrt{5}>1\)

nên \(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}>1\)

b: \(0< \dfrac{1}{5}< 1\)

=>\(\left(\dfrac{1}{5}\right)^{\sqrt{2}}< \left(\dfrac{1}{5}\right)^0=1\)