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kak16kak17
@kak16kak17
June 2021
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Нужна помощь) Пожалуйста ))))
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Vas61
㏒4(2sinπ/6)=㏒4(2×1/2)=㏒4(1)=0 т.к 4^0=1
㏒3(74)/㏒27(74)=1/㏒74(3):1/㏒74(27)=㏒74(27)/㏒74(3)=㏒3(27)=3 т.к.3³=27
㏒2(27)/㏒16(27)=1/㏒27(2):1/㏒27(16)=㏒27(16)/㏒27(2)=㏒2(16)=4 т.к.2^4=16
㏒7(2cos⇅/3)=㏒7(2×1/2)=㏒7(1)=0 т.к. 7^0=1
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Answers & Comments
㏒3(74)/㏒27(74)=1/㏒74(3):1/㏒74(27)=㏒74(27)/㏒74(3)=㏒3(27)=3 т.к.3³=27
㏒2(27)/㏒16(27)=1/㏒27(2):1/㏒27(16)=㏒27(16)/㏒27(2)=㏒2(16)=4 т.к.2^4=16
㏒7(2cos⇅/3)=㏒7(2×1/2)=㏒7(1)=0 т.к. 7^0=1