zadanie 1.4 zadanie w załączniku PILNE!

Odpowiedzi w załącznikach :)) 

[latex]a)\(frac{2}{3}-2^{-2})^{-1}=(frac{2}{3}-frac{1}{2}^{2})^{-1}=(frac{2}{3}-frac{1}{4})^{-1}=\\(frac{8}{12}-frac{3}{12})^{-1}=(frac{5}{12})^{-1}=(frac{12}{5})^{1}=frac{12}{5}=2frac{2}{5}[/latex] [latex]b)\( sqrt[3]{8^{-1}} + sqrt[4]{9^{-2}} )^{-1}=( sqrt[3]{frac{1}{8}^{1}} + sqrt[4]{frac{1}{9}^{2}} )^{-1}=\\( sqrt[3]{frac{1}{8}}} + sqrt[4]{frac{1}{81}} )^{-1}=(frac{1}{2}+frac{1}{3})^{-1}=(frac{3}{6}+frac{2}{6})^{-1}=(frac{5}{6})^{-1}=(frac{6}{5})^{1}=frac{6}{5}=1frac{1}{5}[/latex] [latex]c)\((frac{3}{4})^{-1}-(1,5)^{-1})^{-2}=(frac{4}{3}-(frac{3}{2})^{-1})^{-2}=(frac{4}{3}-frac{2}{3})^{-2}=\\(frac{2}{3})^{-2}=(frac{3}{2})^{2}=frac{9}{4}=2frac{1}{4}[/latex] [latex]d)\((1,25)^{-1}- sqrt[3]{frac{8}{27}} )^{-2}=((1frac{1}{4})^{-1}-frac{2}{3})^{-2}=((frac{5}{4})^{-1}-frac{2}{3})^{-2}=(frac{4}{5}-frac{2}{3})^{-2}=\(frac{12}{15}-frac{10}{15})^{-2}=(frac{2}{15})^{-2}=(frac{15}{2})^{2}=frac{225}{4}=56frac{1}{4}[/latex]