Odpowiedź:
[tex]Zad\ \ 1.\\\\a)\ \ 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\\\\b)\ \ \sqrt{64}-\sqrt[3]{-125}=8-(-\sqrt[3]{125})=8-(-5)=8+5=13\\\\c)\ \ \sqrt{100}-\sqrt{36}+\sqrt{100-36}=10-6+\sqrt{64}=4+8=12\\\\d)\ \ \sqrt{900}:\sqrt[3]{8}+\sqrt{81}=30:2+9=15+9=24[/tex]
[tex]Zad\ \ 2.\\\\a)\ \ \dfrac{2^6+2^5}{12}=\dfrac{2\cdot2^6+2^5}{12}=\dfrac{(2+1)\cdot2^5}{12}=\dfrac{\not3^1\cdot2^5}{\not12_{4}}=\dfrac{2^5}{4}=\dfrac{2^5}{2^2}=2^{5-2}=2^3=8\\\\lub\\\\\dfrac{2^6+2^5}{12}=\dfrac{64+32}{12}=\dfrac{96}{12}=8 \\\\\\b)\ \ \dfrac{8^2\cdot3^6}{6^5}=\dfrac{(2^3)^2\cdot3^6}{2^5\cdot3^5}=\dfrac{2^6\cdot3^6}{2^5\cdot3^5}=2^{6-5}\cdot3^{6-5}=2^1\cdot3^1=2\cdot3=6[/tex]
