Question

A expressão $x= \frac{1}{ \sqrt{3}+ \sqrt{2} } + \frac{2}{ \sqrt{2} }$ é igual a:

a)$\frac{2 \sqrt{3} }{3}$
b)$\frac{3 \sqrt{3} }{2}$
c)0
d)$\sqrt{3}$
e)$2^{ \sqrt{3} }$

Answer 1

Fica assim:

$\dfrac{1}{ \sqrt{3 + \sqrt{2}}} + \dfrac{2}{ \sqrt{2}}$

No primeiro termo inverte a fração e muda o sinal, no segundo termo fica √2

$- \sqrt{2} + \sqrt{3} + \sqrt{2} \\ \\ \\ => \sqrt{3}$



Resposta letra d) √3

Answer 2

$x= \frac{ \sqrt{2}+2( \sqrt{3}+ \sqrt{2}) }{ \sqrt{2} ( \sqrt{3}+ \sqrt{2}) }$

$x= \frac{ \sqrt{2}+2 \sqrt{3}+ 2\sqrt{2}) }{ \sqrt{6}+ 2 } = \frac{3 \sqrt{2}+2 \sqrt{3} }{ \sqrt{6}+2 }$

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

$x=\frac{(3 \sqrt{2}+2 \sqrt{3})( \sqrt{6}-2) }{2} =\frac{3 \sqrt{12}-6 \sqrt{2}+2 \sqrt{18} -4 \sqrt{3} }{2}$

$x=\frac{3 \sqrt{4.3}-6 \sqrt{2}+2 \sqrt{9.2} -4 \sqrt{3} }{2} =\frac{6 \sqrt{3}-6 \sqrt{2}+6 \sqrt{2} -4 \sqrt{3} }{2}$

$x=3 \sqrt{3}-3 \sqrt{2}+3 \sqrt{2} -2 \sqrt{3}$

$x=(3 \sqrt{3}-2 \sqrt{3})+(3 \sqrt{2} -3 \sqrt{2})$

$x= \sqrt{3}+0$

$x= \sqrt{3}$