Question

Me ajude na questão 3 por favor?

Answer 1

A)

$\frac{1}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{ \sqrt{3} }{ \sqrt{9} } = \frac{ \sqrt{3} }{3}$[/tex]

B)

$\frac{3}{10 \sqrt{5} } \times \frac{ \sqrt{5} }{ \sqrt{5} } = \frac{3 \sqrt{5} }{10 \sqrt{25} } = \frac{3 \sqrt{5} }{50}$

C)

$\frac{10}{2 - \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{10 \sqrt{2} }{2 - \sqrt{4} } = \frac{10 \sqrt{2} }{2 - 2} = 10 \sqrt{2}$

D)

$\frac{2 - \sqrt{2} }{ \sqrt{2 } - 1} \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{2 - \sqrt{4} }{ \sqrt{4 } - 1 } = \frac{2 - 2}{2 - 1} = \frac{0}{1} = 0$

Answer 2

Resposta:

Explicação passo-a-passo:

$a)\frac{1}{\sqrt3}.\frac{\sqrt3}{\sqrt3}=\frac{\sqrt3}{\sqrt9}=\frac{\sqrt3}{3}\\\\b)\frac{3}{10\sqrt5}.\frac{\sqrt5}{\sqrt5}=\frac{3\sqrt5}{10\sqrt25}=\frac{3\sqrt5}{10.5}=\frac{3\sqrt5}{50}\\\\c)\frac{10}{2-\sqrt2}.\frac{2+\sqrt2}{2+\sqrt2}=\frac{20+10\sqrt2}{4+2\sqrt2-2\sqrt2-\sqrt4}=\frac{20+10\sqrt2}{4-2}=\frac{20+10\sqrt2}{2}=10+5\sqrt2\\\\d)\frac{2-\sqrt2}{\sqrt2-1}.\frac{\sqrt2+1}{\sqrt2+1}=\frac{2\sqrt2+2-\sqrt4-\sqrt2}{\sqrt4+\sqrt2-\sqrt2-1}=\frac{2-2+2\sqrt2-\sqrt2}{2-1}=\sqrt2$