Question

Valendo 25 pontos, me ajudem

Answer 1

7) 
Aplicando a distributiva:

a) √3(√18 + √6) = √54 + √18 = 3√6 + 3√2 = 3(√6 + √2)

b) (√10 - 2).(√2 + √5)= (√20 + √50 - 2√2 - 2√5)= (2√5 + 5√2 - 2√2 - 2√5)= 3√2

obs: fatorando √20 = 2√5  e √50= 5√2

c) (√6 - √3)² = (√6)² - 2√6√3 + (√3)² =  6 - 2√18 + 3 = 9 - 6√2

d) (2 + √10)  =  2(1 + √10/2)

8) 
a) $\frac{6}{( \sqrt{6} - \sqrt{2} )} . \frac{( \sqrt{6}+ \sqrt{2} ) }{( \sqrt{6}+ \sqrt{2}) } = \frac{6.( \sqrt{6} + \sqrt{2}) }{ (\sqrt{6})^2 - (\sqrt{2} )^2} = \frac{6 (\sqrt{6} + \sqrt{2}) }{6-2} = \frac{6 (\sqrt{6}+ \sqrt{2}) }{4} = \frac{3( \sqrt{6}+ \sqrt{2}) }{2}$

b)$\frac{ (\sqrt{2} - \sqrt{3}) }{( \sqrt{2}+ \sqrt{3}) } . \frac{( \sqrt{2} - \sqrt{3}) }{ (\sqrt{2} - \sqrt{3}) } = \\ \\ \frac{ (\sqrt{2})^2-2 \sqrt{2} \sqrt{3} + (\sqrt{3})^2}{ (\sqrt{2})^2 - ( \sqrt{3})^2} = \frac{2 - 2 \sqrt{6} + 3}{2-3} = - (5 + 2 \sqrt{6})$

c) $\frac{2}{ \sqrt[5]{4} } . \frac{ \sqrt[5]{4^4} }{ \sqrt[5]{4^4} } = \frac{2 \sqrt[5]{4^4} }{ \sqrt[5]{4^5} } = \frac{2 \sqrt[5]{4^4} }{4} = \frac{ \sqrt[5]{4^4} }{2}$

d) $\frac{3}{8 \sqrt[4]{2^3} } . \frac{ \sqrt[4]{2} }{ \sqrt[4]{2}} = \frac{3 \sqrt[4]{2} }{8 \sqrt[4]{2^4} } = \frac{3 \sqrt[4]{2} }{8.2} = \frac{3 \sqrt[4]{2} }{16}$


e)$\frac{12}{ (3+\sqrt{5}) }. (\frac{3- \sqrt{5}) }{(3- \sqrt{5}) } = \frac{12(3- \sqrt{5}) }{3^2-( \sqrt{5})^2 } = \frac{12( 3-\sqrt{5}) }{9-5} = \frac{12 (3-\sqrt{5}) } {4} = 3( 3-\sqrt{5})$

Answer 2

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