Potrzebne na szybkoo !!!!.... Question from @Linox28

Linox28 @Linox28

January 2019 1 8 Report

Potrzebne na szybkoo !!!!

unicorn05 5.
$(3\frac38)^{-\frac13}=(\frac{27}8)^{-\frac13}=(\frac8{27})^{\frac13}=\sqrt[3]{\frac8{27}}=\frac{\sqrt[3]8}{\sqrt[3]{27}}=\frac23$

6.
$a)\\\\9\cdot\sqrt[3]{81}=3^2\cdot81^{\frac13}=3^2\cdot(3^4)^{\frac13}=3^{2+4\cdot\frac13}=3^{3\frac13}\\\\b)\\\\ \left(\sqrt[3]{16}\cdot4^{-2}\right )^3=\left(\sqrt[3]{2^4}\right )^3\cdot\left((2^2)^{-2}\right )^3=2^4\cdot2^{-12}=2^{-8}$

7.
$a)\\\log_3\frac1{18}-\log_3\frac12=\log_3(\frac1{18}:\frac12)=\log_3(\frac1{18}\cdot\frac21)=\log_3\frac19=\\\\=\log_3(\frac13)^2=\log_3 3^{-2}=-2\log_33=-2\\\\ b)\\\log_x 9=-2\ \ \iff\ \ x^{-2}=9\ \ \ \ \ /^{-\frac12}\\~\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{-2\cdot(-\frac12)}=9^{-\frac12}\\~\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x =(3^2)^{-\frac12}= 3^{-1}=\frac13\\\\c)\\x=\log_5 5-\log_5 125=1-\log_5 5^3=1-3\log_55=1-3=-2$

$d)\\x=\log_327-\log_31=\log_33^3-0=3\log_33=3$

8.
$\dfrac{[5^{-3}\cdot25^{\frac32}\cdot(\frac1{125})^{-\frac23}]^{-0,5}}{0,128^{\frac32}:3,2^{\frac32}}=\dfrac{[5^{-3}\cdot(5^2)^{\frac32}\cdot125^{\frac23}]^{-0,5}}{(0,128:3,2)^{\frac32}}=\\\\\\=\dfrac{[5^{-3}\cdot5^3\cdot(5^3)^{\frac23}]^{-0,5}}{(0,04)^{\frac32}}=\dfrac{[5^{-3+3}\cdot5^{2}]^{-0,5}}{(0,04^{\frac12})^3}=\dfrac{(5^0\cdot5^{2})^{-0,5}}{(\sqrt{0,04})^3}=$

$=\dfrac{(1\cdot5^{2})^{-0,5}}{0,2^3}=\dfrac{5^{2\cdot(-0,5)}}{0,008}=\dfrac{5^{-1}}{0,008}=\dfrac{\frac15}{0,008}=\dfrac15\cdot\dfrac{1000}8=\dfrac{200}8=25$

9.
$15^{16}-13^{16}=\\\\=(15^{8})^2-(13^{8})^2=\\\\=(15^8-13^8)(15^8+13^8)=\\\\=[(15^4)^2-(13^4)^2](15^8+13^8)=\\\\=(15^4-13^4)(15^4+13^4)(15^8+13^8)=\\\\=[(15^2)^2-(13^2)^2](15^4+13^4)(15^8+13^8)=\\\\=(15^2-13^2)(15^2+13^2)(15^4+13^4)(15^8+13^8)=\\\\=(225-169)(15^2+13^2)(15^4+13^4)(15^8+13^8)=\\\\=56\cdot(15^2+13^2)(15^4+13^4)(15^8+13^8)$

Otrzymaliśmy iloczyn, w którym jednym z czynników jest liczba 56,
więc liczba $15^{16}-13^{16}$ jest podzielna przez 56

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