1273!!! пожалуйста срочно. Created by Truntik. algebra-ru

1273!!! пожалуйста срочно...

July 2022 1 9 Report

1273!!! пожалуйста срочно

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Minsk00 1273. 1) log4(sin(пи/4)); 2) log10(tg(пи/4)); 3) log8(sin(3пи/4))
4) log2(cos(пи/3)); 5) log3(1) -log4(tg(пи/4))*log5(cos(0)).

Решение:
1) log4(sin(пи/4)) = log2²(1/√2) = (1/2)log2(1/2^(1/2))= (1/2)log2(2^(-1/2))=
= (1/2)*(-1/2)log2(2) = -1/4

2) log10(tg(пи/4)) = log10(1) = 0

3) log8(sin(3пи/4)) = log2³(sin(пи/2+пи/4)) =(1/3)log2(cos(пи/4))=
=(1/3)log2(1/√2) = (1/3)log2(1/2^(1/2))= (1/3)log2(2^(-1/2))=(1/3)*(-1/2)log2(2) = -1/6

4) log2(cos(пи/3)) = log2(1/2) = log2(2^(-1)) = -1*log2(2) = -1

5) log3(1) - log4(tg(пи/4))*log5(cos(0)) = 0 - log4(1)*log5(1) = -0*0 = 0

Ответ: 1) -1/4; 2) 0; 3) -1/6; 4) -1; 5) 0.

4 votes Thanks 5

po kakim pravilam opredelyaetsya poryadok vypolneniya operacij v arifmeticheskom vyrab031b8d6501e2cc5aaf6afabcc5932fc 42436

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Truntik August 2022 | 0 Ответы

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