$\sqrt7 = 7^\frac{1}{2};$
$\sqrt[6]{2^5} = 2^\frac{5}{6};$
$\sqrt[4]{243} = 243^\frac{1}{4}= \left(3^5\right)^{\frac{1}{4}} = 3^{\frac{5}{4}};$
$\dfrac{1}{\sqrt2} = 2^{-\frac{1}{2}};$
$\sqrt[7]{\dfrac{1}{125}} = 125^{-\frac{1}{7}} = \left(5^3\right)^{-\frac{1}{7}}= 5^{-\frac{3}{7}}.$