помогите с алгеброй. Created by aruzhanazamatkizi01. algebra-ru

помогите с алгеброй...

aruzhanazamatkizi01 @aruzhanazamatkizi01

July 2022 1 4 Report

помогите с алгеброй

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bertramjeratire

Ответ:

$\frac{ \sqrt[6]{243 \times 3} - 9}{9} = \frac{ \sqrt[6]{729} - 9}{9} = \frac{3 - 9}{9} = - \frac{6}{9} = - \frac{2}{3}$

$\frac{ \sqrt[3]{ \sqrt{x} } \times (\sqrt[3]{ \sqrt[4]{x} }) ^{2} }{ \sqrt[3]{ \sqrt{x} } \times \sqrt[3]{x} } = \frac{( \sqrt[3]{ \sqrt[4]{x} } )^{2}}{ \sqrt[3]{x} } = \frac{ {x}^{ \frac{1}{6} } }{ {x}^{ \frac{1}{3} } } = \frac{1}{ {x}^{ \frac{1}{6} } } = \frac{1}{ \sqrt[6]{x} } = \frac{ \sqrt[6]{ {x}^{5} } }{x}$

$\sqrt{1 + 4 \sqrt{2} + 12 } = \sqrt{(1 + 2 \sqrt{3} )^{2} } = 1 + 2 \sqrt{3}$

4. y'=

$\frac{10 \sqrt[5]{ {x}^{3} } - 3 \sqrt[3]{x} }{5x \sqrt[15]{ {x}^{14} } }$

$\frac{20 {x}^{2} \sqrt[4]{ {x}^{3} } }{11} + \frac{3x \sqrt[3]{ {x}^{2} } }{5} - 5x + c$

1 votes Thanks 1

aruzhanazamatkizi01 спасибооооо большое

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