Odpowiedź:
[tex](\not2^1\cdot\frac{5}{\not8_{4}})^2-(4-\sqrt{2\frac{1}{4}})(1,5\cdot1\frac{2}{3})=(\frac{5}{4})^2-(4-\sqrt{\frac{9}{4}})\cdot(\frac{\not15^5}{\not10_{2}}\cdot\frac{\not5^1}{\not3_{1}})=\frac{25}{16}-(4-\frac{3}{2})\cdot\frac{5}{2}=\\\\=\frac{25}{16}-(\frac{8}{2}-\frac{3}{2})\cdot\frac{5}{2}=\frac{25}{16}-\frac{5}{2}\cdot\frac{5}{2}=\frac{25}{16}-\frac{25}{4}=\frac{25}{16}-\frac{100}{16}=-\frac{75}{16}=-4\frac{11}{16}[/tex]
