Question

Racionar o denominador de cada fração:

a) 1\2+√3

b) 1\√3 - √2

c) 2\3+2√2

d) 6\5-3√2

e) 1\3√2 - √3

Answer 1

Nesses casos, você multiplica numerador e denominador pelo conjugado do denominador, basta apenas trocar o sinal do 2° termo do denominador...


$a)~~ \dfrac{1}{2+ \sqrt{3} } \to~~ \dfrac{1}{2+ \sqrt{3} } * \dfrac{2- \sqrt{3}}{2- \sqrt{3} } \to~~ \dfrac{2- \sqrt{3}}{4-3 } \to~~ \dfrac{2- \sqrt{3}}{1 } \to\\\\\\ \large\boxed{~2- \sqrt{3} ~}$




$b)~~ \dfrac{1}{ \sqrt{3} - \sqrt{2} } \to~~ \dfrac{1}{ \sqrt{3} - \sqrt{2} } * \dfrac{\sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \to~~ \dfrac{\sqrt{3} + \sqrt{2} }{ (\sqrt{3})^2 - ( \sqrt{2})^2 } \to\\\\\\ \dfrac{\sqrt{3} + \sqrt{2} }{ 3-2 } \to~~ \dfrac{\sqrt{3} + \sqrt{2} }{ 1 } \to~~ \large\boxed{~\sqrt{3} + \sqrt{2}~}$




$c)~~ \dfrac{2}{3+ 2 \sqrt{2} } \to~~ \dfrac{2}{3+ 2 \sqrt{2} } * \dfrac{3- 2 \sqrt{2} }{3- 2 \sqrt{2} } \to~~ \dfrac{2(3- 2 \sqrt{2})}{9- 6 \sqrt{2}+6 \sqrt{2}-4 \sqrt{4} } \to\\\\\\ \dfrac{6- 4 \sqrt{2} }{9- 4 \sqrt{4} } \to~~ \dfrac{6- 4 \sqrt{2} }{9- 4*2 } \to~~ \dfrac{6- 4 \sqrt{2} }{9- 8 } \to~~ \dfrac{6- 4 \sqrt{2} }{1} \to~~ \large\boxed{~6- 4 \sqrt{2} ~}$




$d)~~ \dfrac{6}{5-3 \sqrt{2} } \to~~ \dfrac{6}{5-3 \sqrt{2} } * \dfrac{5+3 \sqrt{2}}{5+3 \sqrt{2} } \to~~ \dfrac{6(5+3 \sqrt{2}) }{25-9 \sqrt{4} } \to\\\\\\ \dfrac{30+18 \sqrt{2} }{25-18 } \to~~ \dfrac{30+18 \sqrt{2} }{25-18 } \to~~ \large\boxed{ \dfrac{30+18 \sqrt{2} }{7 }~}$




$e)~~ \dfrac{1}{3 \sqrt{2}- \sqrt{3} } \to~~ \dfrac{1}{3 \sqrt{2}- \sqrt{3} } * \dfrac{3 \sqrt{2}+ \sqrt{3}}{3 \sqrt{2}+ \sqrt{3} } \to\\\\\\ \dfrac{3 \sqrt{2}+ \sqrt{3}}{(3 \sqrt{2})^2- (\sqrt{3})^2 } \to~~ \dfrac{3 \sqrt{2}+ \sqrt{3}}{9*2-3 } \to~~ \dfrac{3 \sqrt{2}+ \sqrt{3}}{18-3 } \to\\\\\\ \large\boxed{~ \dfrac{3 \sqrt{2}+ \sqrt{3}}{15 } ~}$

Answer 2

Racionar o denominador de cada fração:


a) 1\2+√3

b) 1\√3 - √2

c) 2\3+2√2

d) 6\5-3√2

e) 1\3√2 - √3