Question

Racionalizando o denominador da fração $\frac{\sqrt{3} - 1}{2- \sqrt{2} }$ encontramos:

a) $\frac{\sqrt{2} }{2}$
b) $\frac{\sqrt{2} }{4}$
c) $\sqrt{2}$
d) $\frac{\sqrt{6} }{6}$
e) $\frac{2\sqrt{2} }{9}$

Answer 1

Oi

Basta multiplicar por seu ordenado ...

$\dfrac{\sqrt{3} -1}{2-\sqrt{2} } .\dfrac{2+\sqrt{2} }{2+\sqrt{2} } \\\\\\\\\dfrac{(\sqrt{3} -1).(2+\sqrt{2} )}{(2-\sqrt{2} ).(2+\sqrt{2} )} \\\\\\\\\dfrac{2\sqrt{3} +\sqrt{3}.\sqrt{2}-2-\sqrt{2} }{2^{2}-\sqrt{2}^{2} }\\\\\\\\\dfrac{2\sqrt{3}+\sqrt{6} -2-\sqrt{2} }{4-2}$

$\boxed{\dfrac{2\sqrt{3} +\sqrt{6}-2-\sqrt{2} }{2} }$

Bons estudos! :)