Odpowiedź:
[tex]\sqrt{1\frac{15}{49}}\cdot\sqrt{1\frac{24}{25}}-\left(\sqrt{7\frac{1}{9}}-\sqrt{1\frac{13}{36}}}\right)^2:\sqrt{2\frac{1}{4}}=\sqrt{\frac{64}{49}}\cdot\sqrt{\frac{49}{25}}-\left(\sqrt{\frac{64}{9}}-\sqrt{\frac{49}{36}}\right)^2:\sqrt{\frac{9}{4}}=\\\\=\frac{8}{\not7_{1}}\cdot\frac{\not7^1}{5}-(\frac{8}{3}-\frac{7}{6})^2:\frac{3}{2}=\frac{8}{5}-(\frac{16}{6}-\frac{7}{6})^2:\frac{3}{2}=\frac{8}{5}-(\frac{9}{6})^2:\frac{3}{2}=\frac{8}{5}-(\frac{3}{2})^2:\frac{3}{2}=[/tex]
[tex]=\frac{8}{5}-\frac{\not9^3}{\not4_{2}}\cdot\frac{\not2^1}{\not3}=\frac{8}{5}-\frac{3}{2}=\frac{16}{10}-\frac{15}{10}=\frac{1}{10}[/tex]
