DAJĘ 28 PUKNTÓW PLUS NAJ !!! Bardzo Proszę o pomoc daj za to rozwiązanie bardzo dużą ilość punktów !.... Question from @Vanessa596k

October 2018 1 19 Report

DAJĘ 28 PUKNTÓW PLUS NAJ !!! Bardzo Proszę o pomoc daj za to rozwiązanie bardzo dużą ilość punktów !!!

Jednak nie chodzi mi o same wyniki - chciałabym by to zostało rozpisane

Dziękuje

erre

$Zad.1.\\ a) [(0,3)*(0,3)^3]:(0,3)^2=(0,3)^{1+3-2}=(0,3)^2=0,09\\\\ b) [(\frac{2}{3})^3*(\frac{5}{12})^3]:[(\frac{5}{2})^3*(\frac{6}{5})^3]=(\frac{\frac{2}{3}*\frac{5}{12}}{\frac{5}{2}*\frac{6}{5}})^3=(\frac{5}{18}:3)^3=(\frac{5}{18}*\frac{1}{3})^3=\\\\ =(\frac{5}{54})^3\\\\ c) \sqrt[3]{27*8}=\sqrt[3]{27}*\sqrt[3]8=3*2=6\\\\ d) \sqrt3*\sqrt2*\sqrt6=\sqrt{3*2*6}=\sqrt{36}=6\\\\ e) (\sqrt3)^{-5}:(\sqrt{12})^{-5}=(\frac{\sqrt3}{2\sqrt3})^{-5}=(\frac{1}{2})^{-5}=2^{5}=32\\\\$

$f) [(a^{-2})^{-5}]^3:[(a^{-3})^{-1}]^{-2}=a^{-2*(-5)*3}:a^{-3*(-1)*(-2)}=\\\\ =a^{30}:a^{-6}=a^{30-(-6)}=a^{36}\\\\ g) \frac{\sqrt2*\sqrt[3]4*8^{-\frac{1}{3}}}{16^{\frac{1}{2}}*\sqrt[4]2}=\frac{2^{\frac{1}{2}}*2^{\frac{2}{3}}*2^{-1}}{2^2*2^{\frac{1}{4}}}=\frac{2^{\frac{1}{2}+\frac{2}{3}+(-1)}}{2^{2+\frac{1}{4}}}=\frac{2^\frac{1}{6}}{2^{\frac{9}{4}}}=2^{\frac{1}{6}-\frac{9}{4}}=2^{-\frac{25}{12}}\\\\$

$Zad.2.\\\\ a) (\frac{1}{2})^x=16\\\\ (\frac{1}{2})^x=(\frac{1}{2})^{-4}\\\\ x=-4\\\\ b) 10^{1-x}=100\\\\ 10^{1-x}=10^2\\\\ 1-x=2\\\\ x=-1\\\\ c) x^{-2}=81\\\\ x^{-2}=9^2\\\\ x=9$

$Zad.3.\\ a) \sqrt{32}=\sqrt{2*16}=4\sqrt2\\ b) \sqrt[3]{40}=\sqrt[3]{8*5}=2\sqrt[3]5\\ c) \sqrt[3]{24x^6y^2}=\sqrt[3]{8*3x^3x^2y^2}=2x\sqrt[3]{3x^2y^2}\\\\ Zad.4.\\ a) 4\sqrt7=\sqrt{16*7}=\sqrt{112}\\ b) 3\sqrt[3]5=\sqrt[3]{27*5}=\sqrt[3]{135}\\ c) 2x^2\sqrt[3]{3x^3}=\sqrt[3]{8x^6*3x^3}=\sqrt[3]{24x^{18}}$

$Zad.5.\\ a) \frac{2}{3\sqrt8}=\frac{2}{3\sqrt8}*\frac{\sqrt8}{\sqrt8}=\frac{2\sqrt8}{3*8}=\frac{2\sqrt8}{24}=\frac{\sqrt8}{12}\\\\ b) \frac{3\sqrt5}{2\sqrt{12}}= \frac{3\sqrt5}{2\sqrt{12}}*\frac{\sqrt{12}}{\sqrt{12}}=\frac{3\sqrt{60}}{2*12}=\frac{6\sqrt{15}}{24}=\frac{\sqrt{15}}{4}\\\\ Zad.6.\\ a) log_{\frac{1}{2}}x=-1\\ log_{\frac{1}{2}}x=log_{\frac{1}{2}}(\frac{1}{2})^{-1}\\ x=(\frac{1}{2})^{-1}\\ x=2\\\\$

$b) log_{\frac{1}{2}}x=-2\\ log_{\frac{1}{2}}x=log_{\frac{1}{2}}(\frac{1}{2})^{-2}\\ x=(\frac{1}{2})^{-2}\\ x=4\\\\ c) log_{\frac{1}{3}}x=-\frac{1}{2}\\ log_{\frac{1}{3}}x=log_{\frac{1}{3}}(\frac{1}{3})^{-\frac{1}{2}}\\ x=(\frac{1}{3})^{-\frac{1}{2}}\\ x=\sqrt3\\\\ d) log_4x=0\\ log_4x=log_44^0\\ x=4^0\\ x=1\\\\ e) log_x25=2\\ log_x5^2=log_55^2\\ x=5\\\\ f) log_x3=\frac{1}{2}\\ log_x3=log_99^{\frac{1}{2}}\\ log_x3=log_93\\ x=9$

$Zad.7.\\ a) 3^{\frac{1}{2}log_3{16}}=(3^{log_3{16}})^{\frac{1}{2}}=16^{\frac{1}{2}}=\sqrt{16}=4\\\\ b) 3^{2+log_34}=3^2*3^{log_34}=9*4=36\\\\ Zad.8.\\ a) log_2{log100}=log_2{log10^2}=log_22=1\\\\ b) 5^{\frac{log5}{log25}}= 5^{(\frac{log25}{log5})^{-1}}=5^{(log_525)^{-1}}=25^{-1}=\frac{1}{25}$

$Zad.9.\\ a) log_{\frac{1}{2}}8=log_{\frac{1}{2}}(\frac{1}{2})^{-3}=-3\\\\ log_28=log_22^3=3\\\\ log_{16}2=\log_{16}(16)^{\frac{1}{4}}=\frac{1}{4}\\\\ b) log_{\frac{2}{3}}2,25=log_{\frac{2}{3}}\frac{9}{4}=log_{\frac{2}{3}}(\frac{2}{3})^{-2}=-2\\\\ c) log_{\sqrt3}\frac{1}{27}=log_{\sqrt3}(\sqrt3)^{-6}=-6\\\\ d) log_{\sqrt3}16+log_{\sqrt3}0,125-log_{\sqrt3}6=log_{\sqrt3}{16*0,125:6}=\\ =log_{\sqrt3}\frac{1}{3}=log_{\sqrt3}(\sqrt3)^{-2}=-2$