Rozwiaz 1.2 i z 1.4 g i h.... Question from @Princepololo

December 2018 1 10 Report

Rozwiaz 1.2 i z 1.4 g i h

Dariusz93 1.2)

$\sqrt[3]{2\sqrt[3]{2\sqrt[3]{2}}}$ =
$\sqrt[3]{2\sqrt[3]{2 * 2^{\frac{1}{3}}}}$ =
$\sqrt[3]{2\sqrt[3]{2^{\frac{4}{3}}}}$ =
$\sqrt[3]{2 * (2^{\frac{4}{3}})^{\frac{1}{3}}}$ =
$\sqrt[3]{2 * 2^{\frac{4}{9}}}$ =
$\sqrt[3]{2^{\frac{13}{9}}}$ =
$(2^{\frac{13}{9}})^{\frac{1}{3}}$ =
$2^{\frac{13}{27}}$

1.4g)

$[8,25 - 0,5^{-0,5}(2^{-0,5} + 4^{-0,25})]^{\frac{1}{2}}$ =
$[8\frac{1}{4} - \frac{1}{2}^{-\frac{1}{2}}(2^{-\frac{1}{2}} + 4^{-\frac{1}{4}})]^{\frac{1}{2}}$ =
$[\frac{33}{4} - 2^{\frac{1}{2}}((\frac{1}{2})^{\frac{1}{2}} + (\frac{1}{4})^{\frac{1}{4}})]^{\frac{1}{2}}$ =
$[\frac{33}{4} - 2^{\frac{1}{2}}((\frac{1}{2})^{\frac{1}{2}} + ((\frac{1}{2})^2)^{\frac{1}{4}})]^{\frac{1}{2}}$ =
$[\frac{33}{4} - 2^{\frac{1}{2}}((\frac{1}{2})^{\frac{1}{2}} + (\frac{1}{2})^{\frac{1}{2}})]^{\frac{1}{2}}$ =
$\sqrt{\frac{33}{4} - \sqrt{2}(\sqrt{\frac{1}{2}} + \sqrt{\frac{1}{2}})}$ =
$\sqrt{\frac{33}{4} - \sqrt{2} * 2 * \sqrt{\frac{1}{2}}}$ =
$\sqrt{\frac{33}{4} - 2}$ =
$\sqrt{\frac{33}{4} - \frac{8}{4}}$ =
$\sqrt{\frac{25}{4}}$ = $\frac{5}{2}$ =
$2\frac{1}{2}$

1.4h)

$[12\frac{5}{8} + (\frac{1}{3})^{-2} * (2 * 3^{-1} - 9^{-\frac{1}{2}})]^{-\frac{1}{3}}$ =
$[\frac{101}{8} + 3^2 * (2 * \frac{1}{3} - (\frac{1}{9})^{\frac{1}{2}})]^{-\frac{1}{3}}$ =
$[\frac{101}{8} + 9 * (2 * \frac{1}{3} - ((\frac{1}{3})^2)^{\frac{1}{2}})]^{-\frac{1}{3}}$ =
$[\frac{101}{8} + 9 * (2 * \frac{1}{3} - \frac{1}{3})]^{-\frac{1}{3}}$ =
$[\frac{101}{8} + 9 * \frac{1}{3}(2 - 1)]^{-\frac{1}{3}}$ =
$[\frac{101}{8} + 9 * \frac{1}{3}]^{-\frac{1}{3}}$ =
$[\frac{101}{8} + 3]^{-\frac{1}{3}}$ =
$[\frac{101}{8} + \frac{24}{8}]^{-\frac{1}{3}}$ =
$[\frac{125}{8}]^{-\frac{1}{3}}$ =
$[\frac{8}{125}]^{\frac{1}{3}}$ =
$\sqrt[3]{\frac{8}{125}}$ =
$\frac{2}{5}$

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