Question

Calculand
$( \frac{ 2\sqrt{3 }+ \sqrt{2} }{ 3\sqrt{2} - \sqrt{3} } - \frac{ 2\sqrt{3}- \sqrt{2} }{4 \sqrt{3} + 3\sqrt{2} } ): \frac{19}{ 4\sqrt{6}+1 }$
,se obtine............

Answer 1

$\frac{(2 \sqrt{3} + \sqrt{2})( 3 \sqrt{2}+ \sqrt{3}) }{(3 \sqrt{2} - \sqrt{3})(3 \sqrt{2} + \sqrt{3}) } - \frac{(2 \sqrt{3}- \sqrt{2})(4 \sqrt{3} -3 \sqrt{2} ) }{(4 \sqrt{3} +3 \sqrt{2})(4 \sqrt{3}-3 \sqrt{2}) } * \frac{4 \sqrt{6}+1 }{19} =$
$( \frac{7 \sqrt{6}+12 }{15} - \frac{30-10 \sqrt{6} }{30} )* \frac{4 \sqrt{6} +1}{19} =$
$( \frac{2(7 \sqrt{6} +12)-30+10 \sqrt{6} }{30}* \frac{4 \sqrt{6}+1 }{19} =$
$\frac{14 \sqrt{6}+24-30+10 \sqrt{6} }{30} * \frac{4 \sqrt{6} +1}{19} =$
$\frac{24 \sqrt{6}-6 }{30} * \frac{4 \sqrt{6}+1 }{19} =$
$\frac{6(4 \sqrt{6}-1) }{30} * \frac{4 \sqrt{6}+1 }{19} =$
$\frac{4 \sqrt{6} -1}{5} * \frac{4 \sqrt{6}+1 }{19} =$
$\frac{96-1}{95} =1$