un aiutino grazie
un aiutino grazie
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1° riga:
$3^\frac{5}{8} = \sqrt[8]{3^5} = \sqrt[8]{243};$
$4^\frac{2}{3} = \sqrt[3]{4^2} = \sqrt[3]{16};$
$\left(\dfrac{1}{3}\right)^\frac{3}{2} = \sqrt{\left(\frac{1}{3}\right)^3} = \sqrt{\frac{1}{27}};$
$2^{-\frac{4}{3}} = \left(\dfrac{1}{2}\right)^\frac{4}{3} = \sqrt[3]{\left(\frac{1}{2}\right)^4} =\sqrt[3]{\frac{1}{16}} ;$
$\left(\dfrac{1}{4}\right)^{-\frac{4}{3}} = 4^\frac{4}{3} = \sqrt[3]{4^4} = \sqrt[3]{256}.$
2° riga:
$\sqrt7 = 7^\frac{1}{2};$
$\sqrt[6]{2^5} = 2^\frac{5}{6};$
$\sqrt[4]{243} = 243^\frac{1}{4};$
$\dfrac{1}{\sqrt2} = 2^{-\frac{1}{2}};$
$\sqrt[7]{\frac{1}{125}} = 125^{-\frac{1}{7}}.$