jak w zalaczniku !!!!!!! chodzi o zadanie 6
b)
c)
d)
liczę na naj ;)
zad.6]
a]
(2√2)⁸=(2¹×2 do potęgi ½)⁸=(2 do potęgi 3/2)⁸=2¹²
b]
∛8²×∛4=∛64×∛4=4¹×4 do potegi ⅓=4 do potegi 4/3=(2²) do potegi 4/3=2 do potegi8/3
c]
⁵√16×⁵√2=⁵√(16×2)=⁵√32=2¹
d]
2⁶√4³=2×4 do potegi 3/6=2¹×2do potegi½=2 do potegi3/2
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b)![\sqrt[3]{(2^{3})^{2}}= \sqrt[3]{2^{6}}= 2^{\frac{6}{3}}= 2^{2} \sqrt[3]{(2^{3})^{2}}= \sqrt[3]{2^{6}}= 2^{\frac{6}{3}}= 2^{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%282%5E%7B3%7D%29%5E%7B2%7D%7D%3D+%5Csqrt%5B3%5D%7B2%5E%7B6%7D%7D%3D+2%5E%7B%5Cfrac%7B6%7D%7B3%7D%7D%3D+2%5E%7B2%7D)
c)![\sqrt[5]{2^{4}} * \sqrt[5]{2} = 2^{\frac{4}{5}} * 2^{\frac{1}{5}} = 2^{1}=2 \sqrt[5]{2^{4}} * \sqrt[5]{2} = 2^{\frac{4}{5}} * 2^{\frac{1}{5}} = 2^{1}=2](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B2%5E%7B4%7D%7D+%2A+%5Csqrt%5B5%5D%7B2%7D+%3D+2%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D+%2A+2%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D+%3D+2%5E%7B1%7D%3D2)
d)![2* \sqrt[6]{(2^{2})^{3}}= 2^1 * 2^\frac{6}{6}= 2*2= 2^2 2* \sqrt[6]{(2^{2})^{3}}= 2^1 * 2^\frac{6}{6}= 2*2= 2^2](https://tex.z-dn.net/?f=2%2A+%5Csqrt%5B6%5D%7B%282%5E%7B2%7D%29%5E%7B3%7D%7D%3D+2%5E1+%2A+2%5E%5Cfrac%7B6%7D%7B6%7D%3D+2%2A2%3D+2%5E2)
liczę na naj ;)
zad.6]
a]
(2√2)⁸=(2¹×2 do potęgi ½)⁸=(2 do potęgi 3/2)⁸=2¹²
b]
∛8²×∛4=∛64×∛4=4¹×4 do potegi ⅓=4 do potegi 4/3=(2²) do potegi 4/3=2 do potegi8/3
c]
⁵√16×⁵√2=⁵√(16×2)=⁵√32=2¹
d]
2⁶√4³=2×4 do potegi 3/6=2¹×2do potegi½=2 do potegi3/2