1. Ta có
$\dfrac{2\sqrt{7} + 5\sqrt{2}}{\sqrt{5}} = \dfrac{2\sqrt{35} + 5\sqrt{10}}{5}$
$= \dfrac{2}{5} \sqrt{35} + \sqrt{10} < \sqrt{35} + \sqrt{10}$
Vậy $\dfrac{2\sqrt{7} + 5\sqrt{2}}{\sqrt{5}} < \sqrt{35} + \sqrt{10}$.
2. Trục căn thức ta có
$\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}} = \dfrac{(x-3)(\sqrt{x-1} + \sqrt{2})}{x-1-2}$
$= \sqrt{x-1} + \sqrt{2}$
Ta có
$\sqrt{x-1} + \sqrt{2} \geq 0 + \sqrt{2} = \sqrt{2}$
Dấu "=" xảy ra khi $\sqrt{x-1} = 0$ hay $x=1$.