Question

Simplifique oa radicais. A) 5√27 + 9√3 = B)2√9 + 11√3 - 2 = C) √27 - √75 + √28 = D) √5 + √125 = E) - ³√5 - ³√125 - ³√625 = F) 4 ⁵√2 - ⁵√1024 - 2 ⁵√64 =

Answer 1

Explicação passo-a-passo:

a) Veja que:

• $\sf 5\sqrt{27}=5\cdot3\sqrt{3}=15\sqrt{3}$

Assim:

$\sf 5\sqrt{27}+9\sqrt{3}=15\sqrt{3}+9\sqrt{3}$

$\sf 5\sqrt{27}+9\sqrt{3}=(15+9)\sqrt{3}$

$\sf 5\sqrt{27}+9\sqrt{3}=24\sqrt{3}$

b) Note que:

• $\sf 2\sqrt{9}=2\cdot3=6$

Então:

$\sf 2\sqrt{9}+11\sqrt{3}-2=6+11\sqrt{3}-2$

$\sf 2\sqrt{9}+11\sqrt{3}-2=6-2+11\sqrt{3}$

$\sf 2\sqrt{9}+11\sqrt{3}-2=4+11\sqrt{3}$

c) Veja que:

• $\sf \sqrt{27}=3\sqrt{3}$

• $\sf \sqrt{75}=5\sqrt{3}$

• $\sf \sqrt{28}=2\sqrt{7}$

Assim:

$\sf \sqrt{27}-\sqrt{75}+\sqrt{28}=3\sqrt{3}-5\sqrt{3}+2\sqrt{7}$

$\sf \sqrt{27}-\sqrt{75}+\sqrt{28}=-2\sqrt{3}+2\sqrt{7}$

$\sf \sqrt{27}-\sqrt{75}+\sqrt{28}=2\sqrt{7}-2\sqrt{3}$

$\sf \sqrt{27}-\sqrt{75}+\sqrt{28}=2\cdot(\sqrt{7}-\sqrt{3})$

d) Veja que

• $\sqrt{125}=5\sqrt{5}$

Então:

$\sf \sqrt{5}+\sqrt{125}=\sqrt{5}+5\sqrt{5}$

$\sf \sqrt{5}+\sqrt{125}=6\sqrt{5}$

e) Note que:

• $\sf \sqrt[3]{125}=5$

• $\sf \sqrt[3]{625}=5\sqrt[3]{5}$

Então:

$\sf -\sqrt[3]{5}-\sqrt[3]{125}-\sqrt[3]{625}=-\sqrt[3]{5}-5-5\sqrt[3]{5}$

$\sf -\sqrt[3]{5}-\sqrt[3]{125}-\sqrt[3]{625}=-5-6\sqrt[3]{5}$

f) Veja que:

• $\sf \sqrt[5]{1024}=4$

• $\sf 2\sqrt[5]{64}=2\cdot2\sqrt[5]{2}=4\sqrt[5]{2}$

Assim:

$\sf 4\sqrt[5]{2}-\sqrt[5]{1024}-2\sqrt[5]{64}=4\sqrt[5]{2}-4-4\sqrt[5]{2}$

$\sf 4\sqrt[5]{2}-\sqrt[5]{1024}-2\sqrt[5]{64}=-4$

Answer 2

Oie, Td Bom?!

A)

$= 5 \sqrt{27} + 9 \sqrt{3} \\ = 15 \sqrt{3} + 9 \sqrt{3} \\ = (15 + 9) \sqrt{3} \\ = 24 \sqrt{3}$

B)

$= 2 \sqrt{9} + 11 \sqrt{3} - 2 \\ = 2 \: . \: 3 + 11 \sqrt{3} - 2 \\ = 6 + 11 \sqrt{3} - 2 \\ = 4 + 11 \sqrt{3}$

C)

$= \sqrt{27} - \sqrt{75} + \sqrt{28} \\ = 3 \sqrt{3} - 5 \sqrt{3} + 2 \sqrt{7} \\ = - 2 \sqrt{3} + 2 \sqrt{7}$

D)

$= \sqrt{5} + \sqrt{125} \\ = \sqrt{5} + 5 \sqrt{5} \\ = 1 \sqrt{5} + 5 \sqrt{5} \\ = (1 + 5) \sqrt{5} \\ = 6 \sqrt{5}$

E)

$= - \sqrt[3]{5} - \sqrt[3]{125} - \sqrt[3]{625} \\ = - \sqrt[3]{5} - 5 - 5 \sqrt[3]{5} \\ = - 6 \sqrt[3]{5} - 5$

F)

$= 4 \sqrt[5]{2} - \sqrt[5]{1024} - 2 \sqrt[5]{64} \\ = 4 \sqrt[5]{2} - 4 - 4 \sqrt[5]{2} \\ = - 4$

Att. Makaveli1996