Question

Reduza o único radical e simplifique quando for possível

Answer 1

Explicação passo-a-passo:

.

$a) \dfrac{ \sqrt{72} }{ \sqrt{8} }$

$\sqrt{9}$

$3$

.

$b) \dfrac{ \sqrt{5} }{ \sqrt{20} }$

$\dfrac{1}{ \sqrt{4} }$

$\dfrac{1}{2}$

.

$c) \dfrac{ \sqrt{5} \times \sqrt{6} }{ \sqrt{3} }$

$\dfrac{ \sqrt{30} }{ \sqrt{3} }$

$\sqrt{10}$

.

$d) \dfrac{ \sqrt{40} }{ \sqrt{2} \times \sqrt{5} }$

$\dfrac{ \sqrt{40} }{ \sqrt{10} }$

$\sqrt{4}$

$2$

.

$e) \dfrac{ \sqrt{8} }{ \sqrt{20} } \times \dfrac{ \sqrt{10} }{ \sqrt{2} }$

$\dfrac{ \sqrt{80} }{ \sqrt{40} }$

$\sqrt{2}$

Answer 2

Oie, tudo bom?

a)

$= \frac{ \sqrt{72} }{ \sqrt{8} } \\ = \frac{ \sqrt{8} \sqrt{9} }{ \sqrt{8} } \\ = \sqrt{9} \\ = \sqrt{3 {}^{2} } \\ \boxed{ = 3}$

b)

$= \frac{ \sqrt{5} }{ \sqrt{20} } \\ = \frac{ \sqrt{5} }{ \sqrt{5} \sqrt{4} } \\ = \frac{1}{ \sqrt{4} } \\ = \frac{1}{ \sqrt{2 {}^{2} } } \\ = \frac{1}{2} \\ \boxed{ = 0,5}$

c)

$= \frac{ \sqrt{5} \sqrt{6} }{ \sqrt{3} } \\ = \frac{ \sqrt{5 \: . \: 6} }{ \sqrt{3} } \\ = \frac{ \sqrt{30} }{ \sqrt{3} } \\ = \frac{ \sqrt{3} \sqrt{10} }{ \sqrt{3} } \\ = \sqrt{10} \\ \boxed {≈3,16}$

d)

$= \frac{ \sqrt{40} }{ \sqrt{2} \sqrt{5} } \\ = \frac{ \sqrt{40} }{ \sqrt{2 \: . \: 5} } \\ = \frac{ \sqrt{40} }{ \sqrt{10} } \\ = \frac{ \sqrt{10} \sqrt{4} }{ \sqrt{10} } \\ = \sqrt{4} \\ = \sqrt{2 {}^{2} } \\ \boxed { = 2}$

e)

$= \frac{ \sqrt{8} \sqrt{10} }{ \sqrt{20} \sqrt{2} } \\ = \frac{ \sqrt{8 \: . \: 10} }{ \sqrt{20 \: . \: 2} } \\ = \frac{ \sqrt{80} }{ \sqrt{40} } \\ = \frac{ \sqrt{40} \sqrt{2} }{ \sqrt{40} } \\ = \sqrt{2} \\ \boxed {≈1,41}$

Att. NBA YoungBoy