Matematyka - potęgi i pierwiastki

Zadanie 12

Oblicz:

a) \(\displaystyle \sqrt{64} + \sqrt[3]{1000}\)
b) \(\displaystyle \sqrt[3]{7^2 + 15}\)
c) \(\displaystyle \sqrt{13^2} + \sqrt[3]{7^3}\)
d) \(\displaystyle \sqrt{2} \cdot \sqrt{18} + 3 \cdot \sqrt[3]{3} \cdot \sqrt[3]{9}\)
e) \(\displaystyle \frac{\sqrt{75} + 2\sqrt{3}}{\sqrt{3}}\)

RozwiÄ…zanie:

a)

\[\sqrt{64} + \sqrt[3]{1000} = 8 + 10 = 18\]

b)

\[\sqrt[3]{7^2 + 15} = \sqrt[3]{49 + 15} = \sqrt[3]{64} = 4\]

c)

\[\sqrt{13^2} + \sqrt[3]{7^3} = 13 + 7 = 20\]

d)

\[\sqrt{2} \cdot \sqrt{18} + 3 \cdot \sqrt[3]{3} \cdot \sqrt[3]{9}\] \[\sqrt{36} + 3 \cdot \sqrt[3]{27}\] \[6 + 3 \cdot 3 = 6 + 9 = 15\]

e)

\[\frac{\sqrt{75} + 2\sqrt{3}}{\sqrt{3}}\] \[\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\] \[\frac{5\sqrt{3} + 2\sqrt{3}}{\sqrt{3}} = \frac{7\sqrt{3}}{\sqrt{3}} = 7\]
Wyniki zadania 12:

a) \(\displaystyle 18\)
b) \(\displaystyle 4\)
c) \(\displaystyle 20\)
d) \(\displaystyle 15\)
e) \(\displaystyle 7\)