Odpowiedź:
[tex]Zad. 1\\a) \\(-\frac12)^3*(-\frac12)^4*2^7=(-\frac12)^{3+4}*2^7=(-\frac12)^7*2^7=(-\frac12*2)^7=(-1)^7=-1[/tex]
[tex]b)\\0,1^8*0,2^8:0,02^8=(0,1*0,2)^8:0,02^8=0,02^8:0,02^8=(\frac{0.02}{0.02})^8=1^8=1\\c)\\0,5^6*(-0,5)^7:(0,5)^{13}=\frac{0,5^6*(-0.5)^7}{0,5^{13}}=\frac{(-0.5)^7}{0,5^{7}}=(\frac{-0,5}{0,5})^7=(-1)^7=-1\\d)\\\frac{44^4}{22^3}=\frac{(2*22)^4}{22^3}=\frac{2^4*22^4}{22^3}=2^4*22^1=16*22=352\\[/tex]
[tex]e)\\\frac{64^2*36^2}{6^3*2^7}=\frac{(2^6)^2*(6^2)^2}{6^3*2^7}=\frac{2^{12}*6^4}{6^3*2^7}=2^5*6^1=32*6=192\\f)\\\frac{4^6*8^6}{32^5}=\frac{(2^2)^6*(2^3)^6}{(2^5)^5}=\frac{2^{12}*2^{18}}{2^{25}}=\frac{2^{30}}{2^{25}}=2^5=32[/tex]
[tex]Zad. 2\\a)\\\frac{\sqrt{32}+\sqrt2}{\sqrt2}=\frac{\sqrt2(\sqrt{16}+1)}{\sqrt2}=\sqrt{16}+1=4+1=5\\b)\\\frac{\sqrt8+\sqrt{50}}{\sqrt2}=\frac{\sqrt2(\sqrt4+\sqrt{25})}{\sqrt2}=\sqrt4+\sqrt{25}=2+5=7\\c)\\\frac{\sqrt{27}-\sqrt{12}}{\sqrt3}=\frac{\sqrt3(\sqrt9-\sqrt4)}{\sqrt3}=\sqrt9-\sqrt4=3-2=1\\d)\\\frac{2\sqrt{50}+\sqrt{72}}{2\sqrt2}=\frac{\sqrt2(2\sqrt{25}+\sqrt{36})}{2\sqrt2}=\frac{2\sqrt{25}+\sqrt{36}}2=\frac{2*5+6}{2}=\frac{10+6}2=\frac{16}2=8[/tex]
[tex]Zad. 3\\a)\\3\sqrt[3]4*5\sqrt[3]2=15\sqrt[3]{8}=15*2=30\\b)\\\frac{\sqrt[3]{24}}{6\sqrt[3]3}=\frac16*\sqrt[3]{\frac{24}3}=\frac16*\sqrt[3]8=\frac16*2=\frac26=\frac13\\c)\\\frac{\sqrt[3]9*2\sqrt[3]6}{3\sqrt[3]2}=\frac{2\sqrt[3]{54}}{3\sqrt[3]2}=\frac23*\sqrt[3]{\frac{54}2}=\frac23*\sqrt[3]{27}=\frac23*3=2[/tex]
[tex]d)\\3\sqrt{75}:\sqrt3=\frac{3\sqrt{75}}{\sqrt3}=\frac{3\sqrt3*\sqrt{25}}{\sqrt3}=3\sqrt{25}=3*5=15\\e)\\7\sqrt8*\frac{6\sqrt3}{\sqrt6}=7\sqrt8*6*\sqrt{\frac36}=7\sqrt8*6*\sqrt{\frac12}=42\sqrt{8*\frac12}=42\sqrt4=42*2=84\\f)\\\frac{4\sqrt{10}*\sqrt5}{5\sqrt2}=\frac{4\sqrt{50}}{5\sqrt2}=\frac45*\sqrt{\frac{50}2}=\frac45*\sqrt{25}=\frac45*5=4[/tex]
Szczegółowe wyjaśnienie:
[tex]a^x*b^x=(a*b)^7\\a^x:b^x=(\frac{a}b)^x\\a^x*a^y=a^{x+y}\\a^x:a^y=a^{x-y}[/tex]
[tex](a^x)^y=a^{x*y}[/tex]
[tex]\sqrt{a}*\sqrt{b}=\sqrt{a*b}\\\sqrt{a}:\sqrt{b}=\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}b}\\\sqrt[3]a*\sqrt[3]b=\sqrt[3]{a*b}\\\sqrt[3]a:\sqrt[3]b=\frac{\sqrt[3]a}{\sqrt[3]b}=\sqrt[3]{\frac{a}b}[/tex]
[tex](\sqrt[3]a)^3=a\\(\sqrt{a})^2=a[/tex]
