Question

помогите ребята плиз! 1,3,4,6,8,10,12

Answer 1

$1.;frac{sqrt[4]{32}}{sqrt[4]{8}cdotsqrt[4]{64}}=sqrt[4]{frac{32}{8cdot64}}=sqrt[4]{frac1{16}}=pmfrac12\3.;frac{42sqrt[7]{sqrt[18]a}-7sqrt[3]{sqrt[42]a}}{18sqrt[6]{sqrt[21]a}}=frac{42sqrt[126]a-7sqrt[126]a}{18sqrt[126]a}=frac{35sqrt[126]a}{18sqrt[126]a}=frac{35}{18}=1frac{17}{18}$
$4.;y=sqrt[6]{4x^2-3x-7}\4x^2-3x-7geq0\D=9+4cdot4cdot7=9+112=121=(11)^2\x_{1,2}=frac{3pm11}{8}\x_1=-1,;x_2=frac{14}8=1frac34\4(x+1)(x-1frac34)geq0\xin(-infty;;-1]cup[1frac34;;+infty)\6.;8-x=sqrt{13x+10}\O.D.3.:;13x+10geq0Rightarrow xgeq-frac{10}{13}\(8-x)^2=13x+10\64-16x+x^2-13x-10=0\x^2-29x+54=0\D=841-4cdot54=625=(25)^2\x_{1,2}=frac{29pm25}2\x_1=2,;x^2=27\8.;(log_3243)cdot(log_8512)=(log_3(3^5))cdot(log_8(8^3))=5cdot3=15$
$10.;log_{25}0,2=log_{25}frac15=log_{5^2}5^{-1}=-frac12log_55=-frac12\12.;log_{20}400+log_{0,05}20=log_20(20^2)+log_{frac1{20}}20=\=2log_{20}20+log_{20^{-1}}20=2-1=1$