Nếu $\displaystyle {{a}^{\frac{1}{2}}}>{{a}^{\frac{1}{6}}}$và$\displaystyle {{b}^{\sqrt{2}}}>{{b}^{\sqrt{3}}}$thì A. $\displaystyle a<1;0<b<1$ B. $\displaystyle a>1;b<1$ C. $\displaystyle 0<a<1;b<1$ D. $\displaystyle a>1;0<b<1$
Đáp án đúng: D Vì $\left\{ \begin{array}{l}\frac{1}{2}>\frac{1}{6}\\{{a}^{\frac{1}{2}}}>{{a}^{\frac{1}{6}}}\end{array} \right.\Rightarrow a>1$ và$\left\{ \begin{array}{l}\sqrt{2}<\sqrt{3}\\{{b}^{\sqrt{2}}}>{{b}^{\sqrt{3}}}\end{array} \right.\Rightarrow 0<b<1$ Đáp án D