Question

ohh alquem que e muito bom pode resolver isso pra mim vou dar todos os meus pontos mais quero um trabalho bem feito com resultado e as reulosoes....

Answer 1

1 )

a )

$\sqrt{17}. \sqrt{17} = \sqrt{17^2} =17$

b )

$\sqrt[3]{5}. \sqrt[3]{25} = \sqrt[3]{5.25} = \sqrt[3]{125} = \sqrt[3]{5^3}=5$

c )

$\sqrt{2}. \sqrt{40,5} = \sqrt{2*40,5} = \sqrt{81}= \sqrt{9^2}=9$

d )

$\sqrt{7}.3. \sqrt{7} =3 \sqrt{7^2}=3.7=21$

e )

$2 \sqrt{5}.3 \sqrt{5} =6 \sqrt{5^2} =6.5=30$

f )

$\sqrt{3} . \sqrt{6} . \sqrt{6} . \sqrt{3} = \sqrt{3^2}. \sqrt{6^2} =3.6=18$

2 )

$S=b.h\\S=(5 \sqrt{2} )(3 \sqrt{2} )\\S=15 \sqrt{2^2} \\S=15.2\\S=30~cm^2$

3 )

a )

$\boxed{ \frac{ \sqrt{14} }{ \sqrt{2} }~\to~ \sqrt{ \frac{14}{2} } = \sqrt{7} }$

b ) 

$\boxed{ \frac{ \sqrt[3]{20} }{ \sqrt[3]{5} } \to \sqrt[3]{ \frac{20}{5} } \to \sqrt[3]{4} }$

c )

$\boxed{ \frac{8 \sqrt{10} }{ \sqrt{2} } \to~ 8 \sqrt{ \frac{10}{2} } \to~8 \sqrt{5} }$

d )

$\boxed{ \frac{20 \sqrt{20} }{5 \sqrt{2} } \to~ \frac{20}{5} . \sqrt{ \frac{20}{2} } \to~4 \sqrt{10} }$

4 )

$S=c.l\\ \sqrt{195}= \sqrt{15}.l\\ \\l= \frac{ \sqrt{195} }{ \sqrt{15} }$

$L= \sqrt{13} ~cm$

Largura de $\sqrt{13}$ cm

5  )

$S= \frac{(B+b)h}{2}$

$S= \frac{ (\sqrt{50}+ \sqrt{18}) \sqrt{2} }{2}$

$S= \frac{ (\sqrt{50.2}+ \sqrt{18.2}) }{2}$

$S= \frac{ (\sqrt{100}+ \sqrt{36}) }{2}$

$S= \frac{ 10+ 6 }{2}$

$\boxed{S=8~cm^2}$

6 )

a )

$S=( \sqrt{2}+1 )^2\\S= \sqrt{2^2}+2 \sqrt{2} +1\\S=2+ 2\sqrt{3}+1\\S=3+2 \sqrt{2} ~cm^2$

b )

$S=(\sqrt2+\sqrt3)^2\\S= \sqrt{2^2}+2 \sqrt{6} + \sqrt{3^2} \\S=2+2 \sqrt{6}+3\\S=5+2 \sqrt{6}~cm^2$

7 )

a )

$\boxed{ \frac{ \sqrt[3]{40} }{ \sqrt[3]{5} } \to~ \sqrt[3]{ \frac{40}{5} } \to~ \sqrt[3]{8}\to~ \sqrt[3]{2^3}\to~2 }$

b )

$\boxed{ \frac{ \sqrt{490} }{ \sqrt{10} }\to~ \sqrt{ \frac{490}{10} }\to~ \sqrt{49} \to~ \sqrt{7^2}\to~7}$

c )

$\boxed{ \frac{ \sqrt{2}. \sqrt{6} }{ \sqrt{3} } \to~ \frac{ \sqrt{12} }{ \sqrt{3} } . \frac{ \sqrt{3} }{ \sqrt{3} }\to~ \frac{ \sqrt{36} }{ \sqrt{9} } \to~ \frac{4}{3} }$

d )

$\boxed{ \frac{ \sqrt{40} }{ \sqrt{5}. \sqrt{2} } \to~ \frac{ \sqrt{40} }{ \sqrt{10} } \to \sqrt{4} \to~2}$

8 )

a )

$\boxed{ \frac{4+ \sqrt{12} }{2} \to~ \frac{4+2 \sqrt{3} }{2} \to~2+ \sqrt{3} }$

b )

$\boxed{\frac{4- \sqrt{32} }{2} \to~ \frac{4- \sqrt{2^2.2^2.2} }{2}\to~ \frac{4-4 \sqrt{2} }{2} \to~2-2 \sqrt{2} }$


9 )

D )
$\boxed{{\sqrt{8}.( \sqrt{2}+ \sqrt5})}$