90學測數學單選第1題詳解

因為 $c = \left(\dfrac{1}{4}\right)^{\dfrac{1}{4}} = \left(\left(\dfrac{1}{2}\right)^2\right)^{\dfrac{1}{4}} = \left(\dfrac{1}{2}\right)^{\dfrac{2}{4}} = \left(\dfrac{1}{2}\right)^{\dfrac{1}{2}} = a$。 將 $a, b$ 同時換為 $12$ 次方比較大小： $$a^{12} = \left(\left(\dfrac{1}{2}\right)^{\dfrac{1}{2}}\right)^{12} = \left(\dfrac{1}{2}\right)^6 = \dfrac{1}{64}$$ $$b^{12} = \left(\left(\dfrac{1}{3}\right)^{\dfrac{1}{3}}\right)^{12} = \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}$$ 因為 $\dfrac{1}{64} > \dfrac{1}{81}$，所以 $a > b$。故 $a = c > b$。