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mirinka27071999
@mirinka27071999
July 2022
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Срочно! Пожалуйста, помогите с логарифмами в 7 и 10 примерах
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sedinalana
Verified answer
7
log(2)9=a
(a²-2a+2(a+1)²-3a(a+1)+4*(a+1))/(a-2(a+1))=
=(a²-2a+2a²+4a+2-3a²-3a+4a+4)/(a-2a-2)=(3a+6)/(-a-2)=
=3(a+2)/(-(a+2))=-3
10
Перейдем к логарифму по основанию b
(1/3-log(b)√a)/(1-log(b)∛a) +3(log(b)√a+1/2)/(1-log(b)∛a)=
=(1/3-1/2*log(b)a+3/2*log(b)a+3/2)/(1-1/3*log(b)a)=
=(log(b)a+7/6)/(1-1/3*log(b)a)=(2+7/6)/(1-2/3)=19/6*3/1=19/2=9,5
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Answers & Comments
Verified answer
7log(2)9=a
(a²-2a+2(a+1)²-3a(a+1)+4*(a+1))/(a-2(a+1))=
=(a²-2a+2a²+4a+2-3a²-3a+4a+4)/(a-2a-2)=(3a+6)/(-a-2)=
=3(a+2)/(-(a+2))=-3
10
Перейдем к логарифму по основанию b
(1/3-log(b)√a)/(1-log(b)∛a) +3(log(b)√a+1/2)/(1-log(b)∛a)=
=(1/3-1/2*log(b)a+3/2*log(b)a+3/2)/(1-1/3*log(b)a)=
=(log(b)a+7/6)/(1-1/3*log(b)a)=(2+7/6)/(1-2/3)=19/6*3/1=19/2=9,5