Odpowiedź:
[tex]\sqrt{1\frac{9}{16}}\cdot4^2-(\frac{1}{4})^3=\sqrt{\frac{25}{16}}\cdot16-\frac{1}{64}=\frac{5}{\not4_{1}}\cdot\not16^4-\frac{1}{64}=5\cdot4-\frac{1}{64}=20-\frac{1}{64}=\\\\=19\frac{64}{64}-\frac{1}{64}=19\frac{63}{64}\\\\\\\sqrt{1\frac{7}{9}}\cdot3^3-(\frac{1}{3})^2=\sqrt{\frac{16}{9}}\cdot27-\frac{1}{9}=\frac{4}{\not3_{1}}\cdot\not27^9-\frac{1}{9}=4\cdot9-\frac{1}{9}=36-\frac{1}{9}=\\\\=35\frac{9}{9}-\frac{1}{9}=35\frac{8}{9}[/tex]
[tex]\sqrt{1\frac{7}{9}}\cdot3^3-(\frac{1}{4})^2=\sqrt{\frac{16}{9}}\cdot27-\frac{1}{16}=\frac{4}{\not3_{1}}\cdot\not27^9-\frac{1}{16}=4\cdot9-\frac{1}{16}=36-\frac{1}{16}=\\\\=35\frac{16}{16}-\frac{1}{16}=35\frac{15}{16}[/tex]
