1) Ta có
$$2 \sqrt{5} - 5\sqrt{2} = \sqrt{20} - \sqrt{50}$$
Do $20 < 50$ nên $\sqrt{20} - \sqrt{50} < 0 <1$.
Vậy $2\sqrt{5} - 5\sqrt{2} <1$.
2) Ta có
$$\left( \dfrac{\sqrt{8}}{\sqrt{3}}\right)^2 = \dfrac{8}{3} = \dfrac{128}{48}$$
$$\left(\dfrac{3}{4} \right)^2 = \dfrac{9}{16} = \dfrac{27}{48}$$
Ta có $\dfrac{27}{48} < \dfrac{128}{48}$ nên $\left(\dfrac{3}{4} \right)^2 < \left( \dfrac{\sqrt{8}}{\sqrt{3}}\right)^2$.
Vậy $\dfrac{3}{4} < \dfrac{\sqrt{8}}{\sqrt{3}}$.
