Odpowiedź:
[tex]\sqrt{1\frac{9}{16}}\cdot2^3-(\frac{1}{3})^2=\sqrt{\frac{25}{16}}\cdot8-\frac{1}{9}=\frac{5}{\not4_{1}}\cdot{\not8^2}-\frac{1}{9}=10-\frac{1}{9}=9\frac{9}{9}-\frac{1}{9}=9\frac{8}{9}\\\\\\\dfrac{\sqrt{6}\cdot\sqrt{2}}{\sqrt{3}}=\dfrac{\sqrt{12}}{\sqrt{3}}=\sqrt{\frac{12}{3}}=\sqrt{4}=2\\\\\\\sqrt{5\frac{1}{3}}\cdot\sqrt{3}-2^7:2^5=\sqrt{5\frac{1}{3}\cdot3}-2^{7-5}=\sqrt{\frac{16}{\not3}\cdot{\not}3}-2^2=\sqrt{16}-4=4-4=0[/tex]
[tex]\sqrt{12\frac{1}{2}}\cdot\sqrt{3}-3^7:3^4=\sqrt{12\frac{1}{2}\cdot3}-3^{7-4}=\sqrt{\frac{25}{2}\cdot3}-3^3=\sqrt{\frac{75}{2}}-27=\frac{\sqrt{75}}{\sqrt{2}}-27=\\\\=\frac{\sqrt{25\cdot3}}{\sqrt{2}}-27=\frac{5\sqrt{3}}{\sqrt{2}}-27=\frac{5\sqrt{3}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}-27=\frac{5\sqrt{3}\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}}-27=\frac{5\sqrt{6}}{2}-27[/tex]
[tex]2^7:2^4+\sqrt{5}\cdot\sqrt{12\frac{4}{5}}=2^{7-4}+\sqrt{5\cdot12\frac{4}{5}}=2^3+\sqrt{\not5\cdot\frac{64}{\not5}}=8+\sqrt{64}=8+8=16\\\\\\3^6:3^4+\sqrt{2}\cdot\sqrt{12\frac{1}{2}}=3^{6-4}+\sqrt{2\cdot12\frac{1}{2}}=3^2+\sqrt{\not2\cdot\frac{25}{\not2}}=9+\sqrt{25}=9+5=14[/tex]
