Помогите пожалуйста 10-20 балов даю . Created by vladimirklimov12345. algebra-ru

Помогите пожалуйста 10-20 балов даю ...

vladimirklimov12345 @vladimirklimov12345

July 2022 1 1 Report

Помогите пожалуйста 10-20 балов даю

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sangers1959

Verified answer

Объяснение:

№1.

а) 8*2⁴=2³*2⁴=2⁽³⁺⁴⁾=2⁷.

б) 2¹⁰*32=2¹⁰*2⁵=2⁽¹⁰⁺⁵⁾=2¹⁵.

в) 64*16=2⁶*2⁴=2⁽⁶⁺⁴⁾=2¹⁰.

№2.

$\frac{3^5*3^{10}}{3^6*3^7} =\frac{3^{15}}{3^{13}} =3^2=9.\\\frac{2^4*2^6*2^3}{2^5*2^7}=\frac{2^{13}}{2^{12}} =2^1=2.\\\frac{2^6*(2^3)^5}{2^{18}} =\frac{2^6*2^{15}}{2^{18}} =\frac{2^{21}}{2^{18}} =2^3=8.\\\frac{(3^6)^2}{3^6*9} =\frac{3^6}{3^2}=3^4=81.\\\frac{4^7*16}{(4^2)^4} =\frac{4^7*4^2}{4^8} =\frac{4^9}{4^8} =4^1=4.\\\frac{5^6*125}{(5^2)^4} =\frac{5^6*5^3}{5^8}=\frac{5^9}{5^8}=5^1=5.$

№3.

$(\frac{2x^2}{y} )^2=\frac{4x^4}{y^2}.\\ (-\frac{2a^3}{3b})^4= \frac{16a^{12}}{81b^4} .\\(\frac{n^2k^3}{5m})^3=\frac{n^6k^9}{125m^3}.$

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