Dzialania na potegach.
Wykorzystujemy wzory:
[tex]a^b*a^c=a^{b+c}\\a^b:a^c=a^{b-c}\\\frac{a^b}{a^c}=a^b:a^c=a^{b-c}\\a^c*b^c=(a*b)^c\\(a^b)^c=a^{b*c}[/tex]
[tex]a^c:b^c=(\frac{a}b)^c[/tex]
Zad. 1
[tex]a)\\5^6*5^3*5=5^{6+3+1}=5^{10}\\b) \\(\frac12)^4*(\frac12)^2=(\frac12)^{4+2}=(\frac12)^6\\c) \\10^5:10^3=10^{5-3}=10^2\\d) \\\frac{7^8}{7^5}=7^8:7^5=7^{8-5}=7^3\\e)\\(1,5^3)^4=1,5^{3*4}=1,5^{12}\\f)\\10^4*3^4=(10:3)^4=(\frac{10}{3})^4=(3\frac13)^4[/tex]
[tex]g)\\24^3:8^3=(\frac{24}8)^3=3^3\\h) \\\frac{63^2}{9^2}=(\frac{63}{9})^2=7^2[/tex]
Zad. 2
[tex]4^3*2^5=(2*2)^3*2^5=(2^2)^3*2^5=2^6*2^5=2^{6+5}=2^{11}[/tex]
Zad. 3
[tex]125^8:5^{20}=(5^3)^8:5^{20}=5^{24}:5^{20}=5^{24-20}=5^4[/tex]
Pierwiastki:
[tex]\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}[/tex]
[tex]a = \sqrt{a^2} = \sqrt{a*a}\\b = \sqrt[3]{a^3} = \sqrt[3]{a*a*a}[/tex]
Zad. 1
[tex]\sqrt{36}=\sqrt{6*6}=6\\\sqrt{81}= \sqrt{9*9}=9\\\sqrt{1\frac9{16}} = \sqrt{\frac{25}9} = \frac{5}{3}[/tex]
[tex]\sqrt{0,04} = \sqrt{\frac4{100}} = \frac2{10}\\\sqrt[3]{27} = \sqrt[3]{3*3*3}=3\\\sqrt[3]{1000} = \sqrt[3]{10*10*10}= 10\\\sqrt[3]{-\frac18} = \sqrt[3]{(-\frac12)*(-\frac12)*(-\frac12)} = -\frac12\\\sqrt[3]{0,064} = \sqrt[3]{\frac{64}{1000}} = \frac{\sqrt[3]{4*4*4}}{\sqrt[3]{10*10*10}} = \frac4{10}=\frac2{5}[/tex]
Zad. 2
[tex]a) \\\sqrt{64}-7=\sqrt{8*8}-7=8-7=1\\b)\\\sqrt{49} - \sqrt[3]{64}=\sqrt{7*7}-\sqrt[3]{4*4*4}=7-4=3\\c)\\\sqrt[3]{125} - (\sqrt{16}+8) = \sqrt[3]{5*5*5} - (\sqrt{4*4}+8) = 5-(4+8)=5-12=-7\\d) \\\sqrt{900} : \sqrt[3]{27}*2=\sqrt{9*100} : \sqrt[3]{3*3*3}*2=3*10:3*2=30:6=5\\e) \\3\sqrt{25}+\sqrt[3]{-64} = 3\sqrt{5*5}+\sqrt[3]{-4*(-4)*(-4)} = 3*5+(-4)=15-4=11\\f) 8\sqrt[3]{-8}+5\sqrt9=8\sqrt[3]{-2*(-2)*(-2)} + 5\sqrt{3*3} = 8*(-2)+5*3=-16+15=-1[/tex]
