[tex]6)\\\\L= [ ( \frac{1}{4})^{ - \frac{1}{2}} *2^{\frac{1}{9}}]^ {1,8}= [ 4 ^{ \frac{1}{2}} *2^{\frac{1}{9}}]^ {1,8}= [(2^2)^{ \frac{1}{2}} *2^{\frac{1}{9}}]^ {1,8}= [ 2^1*2^{\frac{1}{9}}]^ {\frac{18}{10}}= [ 2^{1+\frac{1}{9}}]^ { \frac{9}{5}}= \\\\=[ 2^{1\frac{1}{9}}]^ { \frac{9}{5}}=[ 2^{ \frac{10}{9}}]^ { \frac{9}{5}}= 2^{ \frac{\not{10}^2}{\not{9}^1}* \frac{\not{9}^1}{\not{5}^1}}= 2^2=4 [/tex]
[tex]P=(\frac{2}{\sqrt{2}}) ^4=\frac{2^4}{\sqrt{2}^4} =\frac{16}{2^2}=\frac{16}{4}=4\\\\L=P [/tex]
[tex]7)\\\\x= \frac{25^{-\frac{1}{3}}*\sqrt[3]{25}*\sqrt[3]{5}} {7^2*(\sqrt[4]{7})^0} =\frac{25^{-\frac{1}{3}+\frac{1}{3} }*\sqrt[3]{5}} {7^2*1} =\frac{25^{0}*\sqrt[3]{5}} {49} =\frac{1*\sqrt[3]{5}}{49}=\frac{ \sqrt[3]{5}}{49}\\\\ Jest\ to\ liczba\ niewymierna,\ poniewaz\ \sqrt[3]{5} \ jest\ liczba\ niewymierna.[/tex]
[tex] Liczba\ niewymierna\ -\ to\ taka\ liczba,\ ktorej\ nie\ mozna\ zapisac\\\\ za\ pomoca\ ulamka\ zwyklego. [/tex]
