so sánh ( $\frac{\pi}{2}$ )^căn 2 và ( $\frac{\pi}{5}$)^-căn3

Đáp án:

$\left(\dfrac{\pi}{5}\right)^{-\sqrt3} > \left(\dfrac{\pi}{2}\right)^{\sqrt2}$

Giải thích các bước giải:

Ta có:

$\left(\dfrac{\pi}{5}\right)^{-\sqrt3}$

$= \left(\dfrac{5}{\pi}\right)^{\sqrt3}$

Xét $\dfrac{5}{\pi} - \dfrac{\pi}{2}$

$= \dfrac{10 - \pi^2}{2\pi}$

$= \dfrac{(\sqrt{10} - \pi)(\sqrt{10} + \pi)}{2\pi}$

Do $\sqrt{10} > \pi$

nên $\dfrac{(\sqrt{10} - \pi)(\sqrt{10} + \pi)}{2\pi} > 0$

$\to \dfrac{5}{\pi} - \dfrac{\pi}{2} > 0$

$\to \dfrac{5}{\pi} >\dfrac{\pi}{2}$

Ta lại có:

$\sqrt3 > \sqrt2$

$\to \left(\dfrac{5}{\pi}\right)^{\sqrt3} > \left(\dfrac{\pi}{2}\right)^{\sqrt2}$

Hay $\left(\dfrac{\pi}{5}\right)^{-\sqrt3} > \left(\dfrac{\pi}{2}\right)^{\sqrt2}$