Oblicz (w załączniku). Jest to zadanie dotyczące potęg o wykładniku wymiernym. Daję naj. ;-).... Question from @Lenkabozenka

December 2018 1 18 Report

Oblicz (w załączniku). Jest to zadanie dotyczące potęg o wykładniku wymiernym. Daję naj. ;-)

ghe A)
$7^{\frac{1}{2}}\cdot7^{\frac{1}{3}}\cdot7^{\frac{1}{6}}=7^{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=$

$7^{\frac{3}{6}+\frac{2}{6}+\frac{1}{6}}=7^{\frac{6}{6}}=7^1=1$
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b)
$(9^{\frac{1}{3}}\cdot9^{\frac{1}{4}}):9^{\frac{1}{12}}=9^{\frac{1}{3}+\frac{1}{4}}-\frac{1}{12}}=$

$9^{\frac{4}{12}+\frac{3}{12}-\frac{1}{12}}=9^{\frac{6}{12}}=9^{\frac{1}{2}}=3$
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c)
$5^{-\frac{1}{2}}:(5^{\frac{1}{4}}\cdot 5^{1\frac{1}{4}})=5^{-\frac{1}{2}}:(5^{\frac{1}{4}+1\frac{1}{4}})=$

$5^{-\frac{1}{2}}:5^{1\frac{2}{4}}=5^{-\frac{1}{2}}:5^{1\frac{1}{2}}=5^{-\frac{1}{2}-1\frac{1}{2}}=5^{-2}=\frac{1}{25}$
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d)
$(16^{-\frac{3}{2}})^{\frac{1}{3}}\cdot(16^{\frac{3}{4}})^{\frac{1}{3}})=16^{-\frac{1}{2}}\cdot16^{\frac{1}{4}}=$

$16^{-\frac{2}{4}+\frac{1}{4}}=16^{-\frac{1}{4}}=\frac{1}{2}$
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e)
$(4^{\frac{3}{4}})^{\frac{2}{3}}:(4^{-\frac{5}{3}})^{\frac{3}{2}}=4^{\frac{1}{2}}:4^{-\frac{5}{2}}=$

$4^{\frac{1}{2}+\frac{5}{2}}=4^{\frac{6}{2}}=4^3=64$
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f)
$(6^{\frac{1}{2}})^{-3}\cdot(6^{\frac{1}{3}})^{-2}\cdot6^{\frac{1}{6}}=6^{-\frac{3}{2}}\cdot6^{-\frac{2}{3}}\cdot6^{\frac{1}{6}}=$

$6^{-\frac{9}{6}}-\frac{4}{6}+\frac{1}{6}}=6^{-\frac{12}{6}}=6^{-2}=\frac{1}{36}$
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g)
$6^{\frac{3}{2}}\cdot 2^{\frac{1}{2}}\cdot 3^{-\frac{1}{2}}= (2\cdot3)^{\frac{3}{2}}\cdot 2^{\frac{1}{2}}\cdot 3^{-\frac{1}{2}}=$

$2^{\frac{3}{2}}\cdot 2^{\frac{1}{2}}\cdot 3^{-\frac{1}{2}}\cdot 3^{\frac{3}{2}}=2^{\frac{3}{2}+\frac{1}{2}}\cdot 3^{-\frac{1}{2}+\frac{3}{2}}=$

$2^{\frac{4}{2}}\cdot 3^{\frac{2}{2}}=2^{2}\cdot 3^{1}=4\cdot 3=12$
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h)
$(15^{\frac{3}{2}}:\sqrt5)\cdot \sqrt3=((3\cdot5)^{\frac{3}{2}}:5^{\frac{1}{2}})\cdot 3^{\frac{1}{2}}=$

$(3^{\frac{3}{2}}\cdot5^{\frac{3}{2}}:5^{\frac{1}{2}})\cdot 3^{\frac{1}{2}}=(3^{\frac{3}{2}}\cdot5^{\frac{3}{2}-\frac{1}{2}})\cdot 3^{\frac{1}{2}}=$

$(3^{\frac{3}{2}}\cdot5^{\frac{2}{2}})\cdot 3^{\frac{1}{2}}=(3^{\frac{3}{2}}\cdot5^1)\cdot 3^{\frac{1}{2}}=$

$3^{\frac{3}{2}}\cdot5\cdot 3^{\frac{1}{2}}=3^{\frac{3}{2}+\frac{1}{2}}\cdot5=$

$3^{\frac{4}{2}}\cdot5=3^2\cdot5=9\cdot5=45$
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i)
$(20^{\frac{3}{4}}\cdot 5^{1\frac{1}{4}}):2^{-\frac{1}{2}}=((4\cdot5)^{\frac{3}{4}}\cdot 5^{1\frac{1}{4}}):2^{-\frac{1}{2}}=$

$(4^{\frac{3}{4}}\cdot 5^{\frac{3}{4}}\cdot 5^{1\frac{1}{4}}:2^{-\frac{1}{2}}=(4^{\frac{3}{4}}\cdot 5^{\frac{3}{4}+1\frac{1}{4}}):2^{-\frac{1}{2}}=$
$(4^{\frac{3}{4}}\cdot 5^{1\frac{4}{4}}):2^{-\frac{1}{2}}=(4^{\frac{3}{4}}\cdot 5^2):2^{-\frac{1}{2}}=$

$((2^2)^{\frac{3}{4}}\cdot 5^2):2^{-\frac{1}{2}}=(2^{\frac{3}{2}}\cdot 5^2):2^{-\frac{1}{2}}=$

$2^{\frac{3}{2}+\frac{1}{2}}\cdot 5^2=2^{\frac{4}{2}}\cdot 5^2=2^2\cdot 5^2=4\cdot25=100$

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