Question

simplify by rationalising the denominator:7√3-5√2/√48+√18

Answer 1

here's the solution. hope it helps.

Answer 2

hiii!!!

here's ur answer...

$given \: = > \frac{7 \sqrt{3} - 5 \sqrt{2} }{ \sqrt{48} - \sqrt{18} } \\ \\ we \: can \: write \: \sqrt{48} = 4 \sqrt{3} and \\ \sqrt{18} = 3 \sqrt{2} \\ \\ = \frac{7 \sqrt{3} - 5 \sqrt{2} }{4 \sqrt{3} - 3\sqrt{2} } \times \frac{4 \sqrt{3} + 3 \sqrt{2} }{4 \sqrt{3} + 3 \sqrt{2} } \\ \\ = \frac{(7 \sqrt{3} - 5 \sqrt{2})(4 \sqrt{3} + 3 \sqrt{2} ) }{(4 \sqrt{3} - 3 \sqrt{2})( 4 \sqrt{3} + 3 \sqrt{2} ) } \\ \\ = \frac{7 \sqrt{3} (4 \sqrt{3} + 3 \sqrt{2} ) - 5 \sqrt{2} (4 \sqrt{3} + 3 \sqrt{2}) }{( {4 \sqrt{3}) }^{2} - ( {3 \sqrt{2} )}^{2} } \\ \\ = \frac{28(3) + 21 \sqrt{6} - 20 \sqrt{6} - 15(2) }{48 - 18} \\ \\ = \frac{84 + \sqrt{6} - 30}{30} \\ \\ = \frac{54 + \sqrt{6} }{30}$

hope this helps..!!