Proszę bardzo o rozwiązanie zaznaczonych zadań z załącznika :DZ góry bardzo dziękuje :D :*.... Question from @ewelcia002

ewelcia002 @ewelcia002

October 2018 1 28 Report

Proszę bardzo o rozwiązanie zaznaczonych zadań z załącznika :D

Z góry bardzo dziękuje :D :*

Epis

z. 3.87.

a) $\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)} - \sqrt{2}=\sqrt{2}+1-\sqrt{2} = 1$

b) $\frac{(\sqrt{3}-2)(\sqrt{3}-2)}{(\sqrt{3}+2)(\sqrt{3}-2)}-\frac{12\sqrt{3}}{3}=\frac{3-2\sqrt{3}-2\sqrt{3}+4}{3-4}-4\sqrt{3}=\frac{7-4\sqrt{3}}{-1}-4\sqrt{3}=4\sqrt{3}-7-4\sqrt{3}=-7$

d) $\frac{3-\sqrt{6}}{(3+\sqrt{6})(3-\sqrt{6})}+\frac{3+\sqrt{6}}{(3-\sqrt{6})(3+\sqrt{6})}=\frac{3-\sqrt{6}}{9-6}+\frac{3+\sqrt{6}}{9-6}=\frac{3-\sqrt{6}+3+\sqrt{6}}{3}=\frac{6}{3}=2$

e) $\frac{(\sqrt{7}+\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}+\frac{(\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}=\frac{7+\sqrt{35}+\sqrt{35}+5}{7-5}+\frac{7-\sqrt{35}-\sqrt{35}+5}{7-5}=\frac{12+2\sqrt{35}}{2}+\frac{12-2\sqrt{35}}{2}=6+\sqrt{35}+6-\sqrt{35} = 12$

f) $\frac{(\sqrt{6}-\sqrt{3})(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}-\frac{(\sqrt{3}+\sqrt{6})(\sqrt{3}+\sqrt{6})}{(\sqrt{3}-\sqrt{6})(\sqrt{3}+\sqrt{6})}=\frac{6-\sqrt{18}-\sqrt{18}+3}{6-3}-\frac{3+\sqrt{18}+\sqrt{18}+6}{3-6}=\frac{9-2\sqrt{18}}{3}-\frac{9+2\sqrt{18}}{-3}=\frac{9-2\sqrt{18}}{3}+\frac{9+2\sqrt{18}}{3}=\frac{18}{3}=6$

z3.89

f) (⁷/₆)⁻³ * (³/₇)⁻³ = (⁶/₇ * ⁷/₃)³ = 2³ = 8

g) (2,8)⁻⁴ * (⁵/₁₄)⁻⁴ = (¹⁰/₂₈)⁴ * (¹⁴/₅)⁴ = (¹⁰/₂₈ * ¹⁴/₅)⁴ = 2⁴ = 16

h) (¹⁶/₃)⁻³* ( ⁷⁵/₁₀₀)⁻³= (³/₁₆*⁴/₃)³ = (¼)³ = ¹/₆₄

z 3.110

a) $\frac{2^{5}*\sqrt[4]{16}:\sqrt[5]{32}}{{(2^{6})^{-\frac{3}{4}}}*\sqrt[3]{8^{5}}}=\frac{2^{5}*2:2}{2^{5\frac{1}{4}}*2^{5}} = 2^{5} : 2^{10\frac{1}{4} =2^{-5\frac{1}{4}$

b) $\frac{12\sqrt{32^{2}}+2^{8}}{2^{\frac{1}{3}}*2^{\frac{2}{3}}+2}=\frac{12*2^{5}+2^{8}}{2+2}=\frac{12*2^{5}+2^{8}}{2^{2}}= 3*2^{3}+2^{6}$

z3.111

a) $\frac{5^{\frac{2}{3}}*5^{-5}:5^{2}*5^{\frac{1}{12}}}{5*5^{\frac{3}{4}}}=\frac{5^{-4\frac{1}{3}}:5^{2}*5^{\frac{1}{12}}}{5^{1\frac{3}{4}}}=\frac{5^{-6\frac{4}{12}}*5^{\frac{1}{12}}}{5^{1\frac{3}{4}}}=5^{-6\frac{3}{12}} : 5^{\frac{5}{4}} = 5^{-6\frac{1}{4}} : 5^{\frac{5}{4}} = 5^{-7\frac{1}{2}$