Đáp án:
$\dfrac{\sqrt{35}}{3\sqrt{2}-2\sqrt{3}} < 3\sqrt{2}+2\sqrt{3}.$
Giải thích các bước giải:
$\dfrac{\sqrt{35}}{3\sqrt{2}-2\sqrt{3}} \text{ và } 3\sqrt{2}+2\sqrt{3}\\ \dfrac{\sqrt{35}}{3\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{35}(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})}\\ =\dfrac{\sqrt{35}(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2})^2-(2\sqrt{3})^2}\\ =\dfrac{\sqrt{35}(3\sqrt{2}+2\sqrt{3})}{6}\\ <\dfrac{\sqrt{36}(3\sqrt{2}+2\sqrt{3})}{6}=\dfrac{6(3\sqrt{2}+2\sqrt{3})}{6}=3\sqrt{2}+2\sqrt{3}$
Vậy $\dfrac{\sqrt{35}}{3\sqrt{2}-2\sqrt{3}} < 3\sqrt{2}+2\sqrt{3}.$
