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