Question

Zadania 5,6,7! Daje naj!

Answer 1

5.

A

[tex] \sqrt[3]{64} + \sqrt[3]{8} = 4 + 2 = 6 = \sqrt[3]{216} [/tex]

B

[tex] \sqrt[3]{64} - \sqrt[3]{8} = 4 - 2 = 2 = \sqrt[3]{8} [/tex]

C

[tex] \sqrt[3]{ - 8} \times \sqrt[3]{8} = - 2 \times 2 = - 4[/tex]

D

[tex] \sqrt[3]{( - 8 {)}^{2} } = \sqrt[3]{64} = 4[/tex]

POPRAWNE ZAPISY: A, D

6.

a)

[tex] \sqrt[3]{a} = \sqrt[3]{8 \times {10}^{15} } = 2 \times {10}^{5} [/tex]

FAŁSZ

b)

[tex] \sqrt[3]{ {a}^{2} } = \sqrt[3]{(8 \times {10}^{15} {)}^{2} } = \sqrt[3]{8 \times {10}^{15} {}}^{2} = (2 \times {10}^{5} {)}^{2} = {2}^{2} \times {10}^{10} = 4 \times {10}^{10} [/tex]

Wykorzystałem

[tex]\boxed{\large{ \sqrt[n]{ {a}^{m} } = \sqrt[n]{a {}}^{m} }}[/tex]

PRAWDA

7.

[tex] \sqrt[3]{500} = \sqrt[3]{125 \times 4} = 5\sqrt[3]{4} [/tex]

[tex] \sqrt[3]{4} \approx1.58[/tex]

Nasza liczba w przybliżeniu wynosi:

[tex]5 \sqrt[3]{4} = 5 \times 1.58 = 7.9[/tex]

Zatem

[tex] \large{\boxed{7 < \sqrt[3]{500} < 8}}[/tex]