Question

racionaliza os denominadores
a) 1 √3
b) 3 2√3
c) √3 4√2
d) 1 2+√2
e) √3 √2- 2
f) 2 7+√2

Answer 1

RESOLUÇÃO:

a)

$\frac{1}{ \sqrt{3} } = \frac{1 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } = \frac{ \sqrt{3} }{ ({ \sqrt{3} })^{2} } = \frac{ \sqrt{3} }{3}$

b)

$\frac{3}{2 \sqrt{3} } = \frac{3 \sqrt{3} }{2 \times \sqrt{3} \times \sqrt{3} } = \frac{3 \sqrt{3} }{2 \times 3} = \frac{ \sqrt{3} }{2}$

c)

$\frac{ \sqrt{3} }{4 \sqrt{2} } = \frac{ \sqrt{3} \times \sqrt{2} }{4 \sqrt{2} \times \sqrt{2} } = \frac{ \sqrt{3 \times 2} }{4 \times 2} = \frac{ \sqrt{6} }{8}$

d)

$\frac{1}{2 + \sqrt{2} } = \frac{2 - \sqrt{2} }{(2 + \sqrt{2})(2 - \sqrt{2}) } = \frac{2 - \sqrt{2} }{{2}^{2} - ({ \sqrt{2} })^{2} } = \frac{2 - \sqrt{2} }{4 - 2} = \frac{2 - \sqrt{2} }{2}$

e)

$\frac{ \sqrt{3} }{ \sqrt{2} - 2} = \frac{ \sqrt{3} ( \sqrt{2} + 2)}{( \sqrt{2} - 2)( \sqrt{2} + 2) } = \frac{ \sqrt{3}( \sqrt{2} + 2) }{ ({ \sqrt{2} })^{2} - {2}^{2} } = \frac{ \sqrt{3} ( \sqrt{2} + 2) }{2 - 4} = - \frac{ \sqrt{3}( \sqrt{2} + 2) }{2}$

f)

$\frac{2}{7 + \sqrt{2} } = \frac{2(7 - \sqrt{2}) }{(7 + \sqrt{2})(7 - \sqrt{2}) } = \frac{2(7 - \sqrt{2}) }{49 - 2} = \frac{2(7 - \sqrt{2} )}{47}$

Espero ter ajudado!