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zamaraev1997
@zamaraev1997
July 2022
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ОЧЕНЬ ОЧЕНЬ ОЧЕНЬ СРОЧНО.
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Хильмилли
1) sinα=1/√3; (cosα)^2=1-(sinα)^2=1-1/3=2/3⇒cosα=+(-)√(2/3)
Так как π/2<α<π, то cosα=-√(2/3)
cos(30+α)=cos30*cosα-sin30*sinα=√3/2*(-√(2/3))-1/2*1/√3=-√2/2-1/(2√3)=
=-(√2*√3+1)/(2√3)=-(√6+1)/(2√3)
2) y=(6(a+b)^7*√d)/(5c^4*(m+n)
Формулы:
lg(ab)=lga+lgb; lg(a/b)=lga-lgb; lg(a^m)=mlga
Пусть 6(a+b)^7*√d=p - числитель
5c^4*(m+n)=q - знаменатель
lgy=lg(p/q)=lgp-lgq;
p= 6(a+b)^7*√d= 6(a+b)^7*d^(1/2)⇒
lgp=lg6+lg((a+b)^7)+lg(d^(1/2))=lg6+7*lg(a+b)+1/2*lgd
q=5c^4*(m+n)⇒
lgq=lg5+lg(c^4)+lg(m+n)=lg5+4lgc+lg(m+n)
logy=lg(p/q)=lgp-lgq=(lg6+7*lg(a+b)+1/2*lgd)-(lg5+4lgc+lg(m+n))=
=lg6+7*lg(a+b)+1/2*lgd-lg5-4lgc-lg(m+n)=
=
lg6-lg5+7*lg(a+b)-lg(m+n)+1/2*lgd-4lgc
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Answers & Comments
Так как π/2<α<π, то cosα=-√(2/3)
cos(30+α)=cos30*cosα-sin30*sinα=√3/2*(-√(2/3))-1/2*1/√3=-√2/2-1/(2√3)=
=-(√2*√3+1)/(2√3)=-(√6+1)/(2√3)
2) y=(6(a+b)^7*√d)/(5c^4*(m+n)
Формулы:
lg(ab)=lga+lgb; lg(a/b)=lga-lgb; lg(a^m)=mlga
Пусть 6(a+b)^7*√d=p - числитель
5c^4*(m+n)=q - знаменатель
lgy=lg(p/q)=lgp-lgq;
p= 6(a+b)^7*√d= 6(a+b)^7*d^(1/2)⇒
lgp=lg6+lg((a+b)^7)+lg(d^(1/2))=lg6+7*lg(a+b)+1/2*lgd
q=5c^4*(m+n)⇒
lgq=lg5+lg(c^4)+lg(m+n)=lg5+4lgc+lg(m+n)
logy=lg(p/q)=lgp-lgq=(lg6+7*lg(a+b)+1/2*lgd)-(lg5+4lgc+lg(m+n))=
=lg6+7*lg(a+b)+1/2*lgd-lg5-4lgc-lg(m+n)=
=lg6-lg5+7*lg(a+b)-lg(m+n)+1/2*lgd-4lgc