Question

solve it .....
$\frac{1}{ \sqrt{2} } \div \frac{2}{ \sqrt{3} } + 2$
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Answer 1

${\underline{\underline{\huge{\mathfrak{\green{Answer:-}}}}}}$

Hey Mate!!

$\frac{1}{\sqrt{2} } \div \frac{2}{\sqrt{3} }+2\\\\=> \frac{1}{\sqrt{2} } \times \frac{\sqrt{3}}{2 }+2\\\\=> \frac{\sqrt{3} }{2\sqrt{2} } +2\\\\=> \frac{\sqrt{3}+4\sqrt{2} }{2\sqrt{2} } \\\\=> \frac{(\sqrt{3}+4\sqrt{2})(2\sqrt{2}) }{(2\sqrt{2} )(2\sqrt{2}) }\\\\=> \frac{2\sqrt{6} +16 }{8} \\\\=> \frac{2(\sqrt{6} +8) }{8}\\\\=> \frac{\sqrt{6} +8 }{4}\\\\=> \frac{\sqrt{6} }{4} + 2$