[tex]\huge\begin{array}{ccc}a)&2^{\frac{11}{2}}\\b)&2^5\end{array}[/tex]
Definicja potęgi o wykładniku
Twierdzenia:
[tex]a^n\cdot a^m=a^{n+m}\\\\\dfrac{a^n}{a^m}=a^{n-m}\qquad\text{dla}\ a\neq0\\\\\left(a^n\right)^m=a^{n\cdot m}[/tex]
[tex]\sqrt[n]{a^n}=a\qquad\text{dla}\ a\geq0[/tex]
a)
[tex]\dfrac{4^3\cdot16^{\frac{1}{4}}:\sqrt[5]{32}}{64^{-\frac{3}{4}}\cdot8^{\frac{5}{3}}}[/tex]
Zamieniamy wszystkie liczby na potęgi liczby 2:
[tex]=\dfrac{\left(2^2\right)^3\cdot\left(2^4\right)^{\frac{1}{4}}:\left(2^5\right)^{\frac{1}{5}}}{\left(2^6\right)^{-\frac{3}{4}}\cdot\left(2^3\right)^{\frac{5}{3}}}[/tex]
Korzystamy z twierdzeń:
[tex]=\dfrac{2^{2\cdot3}\cdot2^{4\cdot\frac{1}{4}}:2^{5\cdot\frac{1}{5}}}{2^{6\cdot\left(-\frac{3}{4}\right)}\cdot2^{3\cdot\frac{5}{3}}}=\dfrac{2^6\cdot2^1:2^1}{2^{-\frac{9}{2}}\cdot2^5}=\dfrac{2^{6+1-1}}{2^{-\frac{9}{2}+\frac{10}{2}}}=\dfrac{2^6}{2^{\frac{1}{2}}}=2^{6-\frac{1}{2}}=\huge\boxed{2^{\frac{11}{2}}}[/tex]
b)
[tex]\dfrac{12\sqrt{32^2}-\left(-\frac{1}{2}\right)^{-8}}{2^{\frac{1}{3}}\cdot4^{\frac{1}{3}}+\left(6^{\frac{1}{2}}\cdot3^{-\frac{1}{2}}\right)^2}[/tex]
Zamieniamy liczby na potęgi liczby 2 lub na iloczyn liczb, z których jedna jest potęgą liczby:
[tex]=\dfrac{3\cdot2^2\cdot2^5-(-2)^8}{2^{\frac{1}{3}}\cdot\left(2^2\right)^{\frac{1}{3}}+\left[\left(2\cdot3\right)^{\frac{1}{2}}\cdot3^{-\frac{1}{2}}\right]^2}=\dfrac{3\cdot2^{2+5}-2^8}{2^{\frac{1}{3}+\frac{2}{3}}+\left(2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\cdot3^{-\frac{1}{2}}\right)^2}\\\\=\dfrac{3\cdot2^7-2^8}{2^1+\left(2^{\frac{1}{2}}\right)^2}=\dfrac{3\cdot2^7-2\cdot2^7}{2+2^{2\cdot\frac{1}{2}}}=\dfrac{2^7\cdot(3-2)}{2+2}=\dfrac{2^7}{4}=\dfrac{2^7}{2^2}=2^{7-2}=\huge\boxed{2^5}[/tex]
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[tex]\huge\begin{array}{ccc}a)&2^{\frac{11}{2}}\\b)&2^5\end{array}[/tex]
Działania na potęgach.
Definicja potęgi o wykładniku
[tex]a^2=a\cdot a\\a^3=a\cdot a\cdot a\\\vdots\\a^n=\underbrace{a\cdot a\cdot...\cdot a}_{n}\\\\\text{ponadto}\ a^1=a\ \wedge\ a^0=1[/tex]
[tex]a^{-n}=\left(\dfrac{1}{a}\right)^n=\dfrac{1}{a^n}[/tex]
[tex]a^{\frac{m}{n}}=\sqrt[n]{a^m}[/tex]
Twierdzenia:
[tex]a^n\cdot a^m=a^{n+m}\\\\\dfrac{a^n}{a^m}=a^{n-m}\qquad\text{dla}\ a\neq0\\\\\left(a^n\right)^m=a^{n\cdot m}[/tex]
[tex]\sqrt[n]{a^n}=a\qquad\text{dla}\ a\geq0[/tex]
ROZWIĄZANIE:
a)
[tex]\dfrac{4^3\cdot16^{\frac{1}{4}}:\sqrt[5]{32}}{64^{-\frac{3}{4}}\cdot8^{\frac{5}{3}}}[/tex]
Zamieniamy wszystkie liczby na potęgi liczby 2:
[tex]=\dfrac{\left(2^2\right)^3\cdot\left(2^4\right)^{\frac{1}{4}}:\left(2^5\right)^{\frac{1}{5}}}{\left(2^6\right)^{-\frac{3}{4}}\cdot\left(2^3\right)^{\frac{5}{3}}}[/tex]
Korzystamy z twierdzeń:
[tex]=\dfrac{2^{2\cdot3}\cdot2^{4\cdot\frac{1}{4}}:2^{5\cdot\frac{1}{5}}}{2^{6\cdot\left(-\frac{3}{4}\right)}\cdot2^{3\cdot\frac{5}{3}}}=\dfrac{2^6\cdot2^1:2^1}{2^{-\frac{9}{2}}\cdot2^5}=\dfrac{2^{6+1-1}}{2^{-\frac{9}{2}+\frac{10}{2}}}=\dfrac{2^6}{2^{\frac{1}{2}}}=2^{6-\frac{1}{2}}=\huge\boxed{2^{\frac{11}{2}}}[/tex]
b)
[tex]\dfrac{12\sqrt{32^2}-\left(-\frac{1}{2}\right)^{-8}}{2^{\frac{1}{3}}\cdot4^{\frac{1}{3}}+\left(6^{\frac{1}{2}}\cdot3^{-\frac{1}{2}}\right)^2}[/tex]
Zamieniamy liczby na potęgi liczby 2 lub na iloczyn liczb, z których jedna jest potęgą liczby:
[tex]=\dfrac{3\cdot2^2\cdot2^5-(-2)^8}{2^{\frac{1}{3}}\cdot\left(2^2\right)^{\frac{1}{3}}+\left[\left(2\cdot3\right)^{\frac{1}{2}}\cdot3^{-\frac{1}{2}}\right]^2}=\dfrac{3\cdot2^{2+5}-2^8}{2^{\frac{1}{3}+\frac{2}{3}}+\left(2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\cdot3^{-\frac{1}{2}}\right)^2}\\\\=\dfrac{3\cdot2^7-2^8}{2^1+\left(2^{\frac{1}{2}}\right)^2}=\dfrac{3\cdot2^7-2\cdot2^7}{2+2^{2\cdot\frac{1}{2}}}=\dfrac{2^7\cdot(3-2)}{2+2}=\dfrac{2^7}{4}=\dfrac{2^7}{2^2}=2^{7-2}=\huge\boxed{2^5}[/tex]