Nếu ${{\left( \sqrt{3}-\sqrt{2} \right)}^{2m-2}}<\sqrt{3}+\sqrt{2}$ thì A. $\displaystyle m>\frac{3}{2}$ B. $\displaystyle m<\frac{1}{2}$ C. $\displaystyle m>\frac{1}{2}$ D. $\displaystyle m e \frac{3}{2}$
Đáp án đúng: C Ta có $\sqrt{3}+\sqrt{2}=\frac{1}{\sqrt{3}-\sqrt{2}}\Rightarrow $${{\left( \sqrt{3}-\sqrt{2} \right)}^{2m-2}}<{{\left( \sqrt{3}-\sqrt{2} \right)}^{-1}}\Leftrightarrow 2m-2>-1\Leftrightarrow m>\frac{1}{2}$ Đáp án C