Добрый день! Решила, надо свериться.. Created by polmenka. algebra-ru

Добрый день! Решила, надо свериться...

polmenka @polmenka

September 2021 1 40 Report

Добрый день!
Решила, надо свериться.

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Regent1828

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$\displaystyle \tt 1).\\\\{} \ \ a). \ \ \Bigg(16-\frac{1}{3}\cdot6^{2}\Bigg)^{3}=\Bigg(16-\frac{36}{3}}\Bigg)^{3}=(16-12)^{3}=4^{3}=64;\\\\\\{} \ \ b). \ \ \left(\frac{3}{4}\right)^{2}\cdot1\frac{1}{3}-(0,5)^{2}=\frac{9}{16}\cdot\frac{4}{3}-0,25=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}=0,5;\\\\\\{} \ \ c). \ \ \frac{1,6}{(0,4)^{2}}-(-3)^{3}=\frac{1,6}{0,16}-(-27)=10+27=37;\\\\\\{} \ \ d). \ \ 3^{4}\cdot\left(-\frac{2}{3\right)^{3}}+\frac{1}{(-1)^{3}}=81\cdot\left(-\frac{8}{27}\right)-1=-24-1=-25;$

$\displaystyle \tt 2).\\\\{} \ \ 1). \ \ x^{12}\cdot x^{10}=x^{12+10}=x^{22}\\\\{} \ \ 2). \ \ x^{18}:x^{13}=x^{18-13}=x^{5}\\\\{} \ \ 3). \ \ (x^{2})^{5}=x^{2\cdot5}=x^{10}\\\\{} \ \ 4). \ \ (xy)^{7}=x^{7}y^{7}\\\\{} \ \ 5). \ \ \left(\frac{x}{3}\right)^{3}=\frac{x^{3}}{27}\\\\\\3).\\\\{} \ \ a). \ \ \frac{7^{9}\cdot7^{11}}{7^{18}}=7^{9+11-18}=7^{2}=49;\\\\{} \ \ b). \ \ \frac{5^{6}\cdot125}{25^{4}}=\frac{5^{6}\cdot5^{3}}{(5^{2})^{4}}=5^{6+3-8}=5;$

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