Odpowiedź:
[tex]c)\ \ \left(\sqrt{\frac{16}{25}}-\sqrt{\frac{5^2}{4^2}}}\right)^2-\left(\frac{1}{2^2\cdot\sqrt{25} }\right)^2=(\frac{4}{5}-\frac{5}{4})^2-(\frac{1}{4\cdot5})^2=(\frac{16}{20}-\frac{25}{20})^2-(\frac{1}{20})^2=\\\\=(-\frac{9}{20})^2-\frac{1}{400}=\frac{81}{400}-\frac{1}{400}=\frac{80}{400}=\frac{1}{5}[/tex]
[tex]d)\ \ \sqrt{6\frac{2}{3}}\cdot\sqrt{15}\cdot(2\sqrt{5})^2-2\cdot(5\sqrt{3})^2=\sqrt{6\frac{2}{3}\cdot15}\cdot4\cdot5-2\cdot25\cdot3=\\\\=\sqrt{\frac{20}{\not3_{1}}\cdot\not15^5}\cdot20-150=\sqrt{20\cdot5}\cdot20-150=\sqrt{100}\cdot20-150=10\cdot20-150=\\\\=200-150=50[/tex]
