Question

ratonalize 5√3+2√5÷5√3-2√5

Answer 1

$\frac{5 \sqrt{3} + 2 \sqrt{5} }{5 \sqrt{3} - 2 \sqrt{5} }$


By Rationalization,


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



We know, (a + b)(a - b) = a² - b²


Applying formula in denominator , we get




$\frac{ {(5 \sqrt{3}) }^{2} + {(2 \sqrt{5} )}^{2} + 2(5 \sqrt{3 } \times 2 \sqrt{5} )}{ {(5 \sqrt{3} )}^{2} - {(2 \sqrt{5}) }^{2} } \\ \\ \\ \\ => \frac{75 + 20 + 20 \sqrt{15} }{75 - 20} \\ \\ \\ \\ \\ \\ => \frac{95 + 20 \sqrt{15} }{55} \\ \\ \\ \\ => \frac{19 + 4 \sqrt{15} }{11}$

Answer 2

Hey friend !!!!!

•°• Here's your answer •°•

$que = \frac{5 \sqrt{3} + 2 \sqrt{5} }{5 \sqrt{3} - 2 \sqrt{5} } \\ \\ = \frac{5 \sqrt{3} + 2 \sqrt{5} }{5 \sqrt{3} - 2 \sqrt{5} } \times \frac{5 \sqrt{3} + 2 \sqrt{5} }{5 \sqrt{3} + 2 \sqrt{5} } \\ \\ now \: \: we \: \: have \: \: to \: \: use \: \: the \: \: identities \: \: here \\ \\ =(a + b)(a + b) = {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ and \\ = (a - b)(a + b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {(5 \sqrt{3} + 2 \sqrt{5}) }^{2} }{ {(5 \sqrt{3}) }^{2} - {(2 \sqrt{5}) }^{2} } \\ \\ = \frac{ {(5 \sqrt{3}) }^{2} + {(2 \sqrt{5}) }^{2} + 2 \times 5 \sqrt{3} \times 2 \sqrt{5} }{75 - 20} \\ \\ = \frac{75 + 20 + 20 \sqrt{15} }{55} \\ \\ = \frac{95 + 20 \sqrt{15} }{55} \\ \\ = \frac{19 + 4 \sqrt{5} }{11}$

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