Question

racionalización
a) 4/ 5√16=
(es raíz quinta de 16)

b) 2√3/2-√3=

c) √5-√2/(√5+√2)=​

Answer 1

Explicación paso a paso:

$\frac{4}{ \sqrt[5]{16} } = \frac{4}{ \sqrt[5]{16} } \times \frac{ \sqrt[5]{( {16})^{4} } }{ \sqrt[5]{ {(16)}^{4} } } \\ = \frac{4 \sqrt[5]{( {16)}^{4} } }{ \sqrt[5]{( {16)}^{5} } } \\ = \frac{4 \sqrt[5]{( {16)}^{4} } }{16} \\ = \frac{ \sqrt[5]{( {16)}^{4} } }{4}$

$\frac{2 \sqrt{3} }{2 - \sqrt{3} } = \frac{(2 \sqrt{3})(2 + \sqrt{3)} }{(2 - \sqrt{3} )(2 + \sqrt{3} )} \\ = \frac{4 \sqrt{3} + 2( { \sqrt{3}) }^{2} }{4 - 3} \\ = \frac{4 \sqrt{3} + 6}{1} \\ = 4 \sqrt{3} + 6$

$\frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } = \frac{( \sqrt{5} - \sqrt{2} ) \times ( \sqrt{5} - \sqrt{2} )}{( \sqrt{5} + \sqrt{2} )( \sqrt{5} - \sqrt{2} )} \\ = \frac{( { \sqrt{5} - \sqrt{2} )}^{2} }{5 - 2} \\ = \frac{( { \sqrt{5} - \sqrt{2} ) }^{2} }{3}$