Proszę o zrobienie tych zadań z obliczeniami (bez podpunktu a) i b) z 2.11) dam naj

[latex]frac{1}{2-sqrt{3}}=frac{2+sqrt{3}}{4-3}=2+sqrt{3}[/latex] [latex][(2-sqrt{3})^{-frac{1}{2}}+(2+sqrt{3})^{-frac{1}{2}}]^2=\=[(2+sqrt{3})^{frac{1}{2}}+(2-sqrt{3})^{frac{1}{2}}]^2=\=3+sqrt{3}+2sqrt{(2+sqrt{3})(2-sqrt{3})}+2-sqrt{3}=4+2sqrt{4-3}=4+2=6in C[/latex] 1. [latex]2^{frac{3}{2}}=2cdotsqrt{2}=frac{2cdot2}{sqrt{2}}=4cdotfrac{1}{sqrt{2}}=4cdotsqrt{frac{1}{2}}\A.[/latex] 2. [latex]sqrt{5sqrt{5}}=(5cdot5^{frac{1}{2}})^{frac{1}{2}}=(5^{frac{3}{2}})^{frac{1}{2}}=5^{frac{3}{4}}\B.[/latex] 3. [latex]10^{frac{4}{5}}approx6,31\10^{-frac{1}{5}}=frac{10^{frac{4}{5}}}{10}approx0,631\C.[/latex] 4. [latex]A.\4^{1frac{3}{5}}=4^{frac{8}{5}}=16^{frac{4}{5}}\FALSZ[/latex] [latex]B.\(2^{frac{1}{2}}+2^{-frac{1}{2}})^2=2+2cdot1+2^{-1}=4frac{1}{2}=frac{9}{2}\FALSZ[/latex] [latex]C.\(2^{frac{1}{2}}-2^{-frac{1}{2}})^2=2-2cdot1+2^{-1}=frac{1}{2}\PRAWDA[/latex] 5.a) [latex]16^{frac{1}{4}}cdot27^{frac{2}{3}}:25^{-0,5}=2cdot9cdotsqrt{25}=18cdot5=90[/latex] b) [latex]sqrt[3]{9}cdotsqrt[5]{27}:sqrt[15]{81}=3^{frac{2}{3}}cdot3^{frac{5}{3}}:3^{frac{4}{15}}=3^{frac{7}{3}-frac{4}{15}}=3^{frac{35-4}{15}}=3^{frac{31}{15}}[/latex] 6. [latex]frac{2^{frac{1}{3}}cdot(frac{1}{2})^{-frac{2}{3}}cdot(frac{1}{4})^{-frac{1}{4}}}{4^{-1}cdotsqrt{8}}=[/latex] [latex]=frac{2^{frac{1}{3}}cdot2^{frac{2}{3}}cdot2^{frac{1}{2}}}{2^{-2}cdot2^{frac{3}{2}}}=frac{2^{frac{3}{2}}}{2^{-frac{1}{2}}}=2^2=4[/latex] 7. [latex]frac{a^{frac{4}{5}}cdotsqrt{a}}{sqrt[3]{a^2}}cdot a^{-0,6}=frac{a^{frac{4}{5}}cdot a^{frac{1}{2}}}{a^{frac{2}{3}}cdot a^{frac{3}{5}}}=frac{a^{frac{13}{10}}}{a^{frac{19}{15}}}=a^{frac{39-38}{30}}=a^{frac{1}{30}}=sqrt[30]{a}[/latex] 2.11. d) [latex]a=frac{2sqrt{2}}{sqrt[3]{2}}=frac{2cdot2^{frac{1}{2}}}{2^{frac{1}{3}}}=2^{frac{3}{2}-frac{1}{3}}=2^{frac{9-2}{6}}=2^{frac{7}{6}}[/latex] [latex]b=sqrt[5]{frac{64}{sqrt{2}}}=(frac{2^6}{2^{frac{1}{2}}})^{frac{1}{5}}=2^{frac{11}{2}cdotfrac{1}{5}}=2^{frac{11}{10}}\frac{7}{6}=frac{35}{30}\frac{11}{10}=frac{33}{30}\a>b[/latex] c) [latex]a=(frac{9sqrt[3]{9}}{sqrt[5]{27}})^4=frac{3^2cdot3^{frac{2}{3}}}{3^{frac{3}{5}}}=3^{frac{8}{3}-frac{3}{5}}=3^{frac{40-9}{15}3^{frac{31}{15}}\b=(sqrt[3]{sqrt[5]{81})^4})^{10}=(((3^{frac{4}{5}})^4)^{frac{1}{3}})^{10}=3^{frac{4}{5}cdot4cdotfrac{1}{3}cdot10}=3^{frac{32}{3}}[/latex] [latex]frac{32}{3}>frac{31}{15}\a

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