Odpowiedź:
[tex]Zad. 1 \text{ Oblicz:}\\a)\\25^{\frac12}=\sqrt{25}=5\\b)\\0.5^{-3}=(\frac12)^{-3}=2^3=8\\c)\\125^{-\frac13}=(\frac{1}{125})^{\frac13}=\sqrt[3]{\frac1{125}}=\frac15\\d)\\(1\frac{11}{25})^{-\frac12}=(\frac{36}{25})^{-\frac12}=(\frac{25}{36})^{\frac12}=\sqrt{\frac{25}{36}}=\frac56\\e)\\(0.49)^{-\frac1{2}}=(\frac{49}{100})^{-\frac12}=(\frac{100}{49})^{\frac12}=\sqrt{\frac{100}{49}}=\frac{10}7=1\frac37\\f)\\log_232 \to 2^{x}=32 \to 2^x=2^5 \to log_232=5[/tex]
[tex]Zad. 2 \text{ Oblicz:}\\a)\\\frac{7^{11}*4+3*7^{11}}{(7^6:49)^2}=\frac{7*7^{11}}{(7^6:7^2)^2}=\frac{7^{12}}{(7^4)^2}=\frac{7^{12}}{7^8}=7^4=2401[/tex]
[tex]b)\\5^{\frac13}*25^{\frac13}-9^{-1}*9^{1\frac12}=5^{\frac13}*(5^2)^{\frac13}-(3^2)^{-1}*(3^2)^{\frac32}=5^{\frac13}*5^{\frac23}-3^{-2}*3^3=5^1-3^1=5-3=2[/tex]
[tex]c)\\(3-\sqrt5)(3+\sqrt5)+(3\sqrt2-\sqrt2)^2=3^2-(\sqrt5)^2+(2\sqrt2)^2=9-5+8=4+8=12[/tex]
[tex]d)\\log_345-log_35=log_3(\frac{45}5)=log_39=2[/tex]
[tex]Zad. 3 \text{ Uprosc wyrazenie:}[/tex]
[tex]a)\\(x-\sqrt5)^2-(x+3\sqrt5)^2=x^2-2\sqrt5x+5-(x^2+6\sqrt5x+45)=x^2-2\sqrt5x+5-x^2-6\sqrt5x-45=-2\sqrt5x-6\sqrt5x+5-45=-8\sqrt5x-40[/tex]
[tex]b)\\(3x-\sqrt2)(3x+\sqrt2)-(3x+\sqrt7)^2=9x^2-2-(9x^2+6x\sqrt7+7)=9x^2-2-9x^2-6x\sqrt7-7=-6x\sqrt7-2-7=-6x\sqrt7-9[/tex]
[tex]Zad. 4 \text{ Przeksztalc wzory:}\\\\a)\\F=\frac{mv^2}{2g} /*2g\\2Fg=mv^2 /:v^2\\m=\frac{2Fg}{v^2}\\\\b)\\\\\frac{1}f=\frac{1}x+\frac{1}y /*f\\1=f(\frac1x+\frac1y)/:(\frac1x+\frac1y)\\f=\frac{1}{\frac1x+\frac1y}\\f=1:(\frac1x+\frac1y)\\f=1:(\frac{y}{xy}+\frac{x}{xy})\\f=1:\frac{y+x}{xy}\\f=1*\frac{xy}{y+x}\\f=\frac{xy}{y+x}[/tex]
Szczegółowe wyjaśnienie:
[tex](a^n)^m=a^{nm}\\a^{-n}=\frac1{a^n}\\a^{\frac1n}=\sqrt[n]{a}\\a^{\frac{n}m}=\sqrt[m]{a^n}\\a^{n}*a^{m}=a^{n+m}\\a^{n}:a^{m}=a^{n-m}\\xa^n+ya^n=(a+y)a^n\\log_ab=c \to a^c=b\\log_ab+log_ac=log_a(b*c)\\log_ab-log_ac=log_a(\frac{b}c)[/tex]
[tex](a-b)(a+b)=a^2-b^2[/tex]
[tex](a+b)^2=a^2+2ab+b^2\\(a-b)^2=a^2-2ab+b^2\\-(-a)=a[/tex]
