Question

Помогите решить со 2 по 8,во 2 нужно упростить

Answer 1

$2.;5^{log_2+log_{25}64}=5^{log_28}cdot5^{log_{25}64}=5^3cdot5^{log_{5^2}64}=5^3cdot5^{frac12log_564}=\=5^3cdot5^{log_5(64)^frac12}=5^3cdot5^{log_58}=5^3cdot8=125cdot8=1000$
$3.;cos3alphacos4alpha+sin3alphasin4alpha-cosleft(alpha-fracpi2right)=\=frac{cosalpha+cos7alpha}{2}+frac{cosalpha-cos7alpha}2-cosleft(fracpi2-alpharight)=cosalpha-sinalpha\4.;4cdot0,5^{x+1}-0,5^x+0,5^{x-2}=20\4cdot0,5cdot0,5^x-0,5^x+0,5^{-2}cdot0,5^x=20\2cdot0,5^x-0,5^x+4cdot0,5^x=20\5cdot0,5^x=20\0,5^x=4\0,5^x=0,5^{-2}\x=-2$
$5.;log_{frac9{16}(2-0,1x)}leqfrac12\O.D.3.:;2-0,1x>0Rightarrow0,1x<2Rightarrowx<20\2log_{frac9{16}}(2-0,1x)leq1\log_{frac34}(2-0,1x)leq1\2-0,1xgeqfrac34\0,1xleqfrac54\xleqfrac{50}4\xleq12,5$
$6.;y=sqrt{left(frac12right)^{2x+9}-frac18}\left(frac12right)^{2x+9}-frac18geq0\left(frac12right)^{2x+9}geqfrac18\left(frac12right)^{2x+9}geqleft(frac12right)^3\2x+9leq3\2xleq-6\xleq-3$
$7.;sinleft(pi+xright)=cosleft(-fracpi3right)\sin(pi+x)=frac12\pi+x=fracpi6+2pi n\x=-frac{5pi}6+2pi n,;ninmathbb{Z}$
$8.;y=sqrt{5x-1}\y'=frac5{2sqrt{5x-1}}\y'(1)=frac5{2sqrt{5-1}}=frac54=1frac14=1,25$

Answer 2

а ты есть в вк??