Question

$\frac{6}{4 - 3 \sqrt{2} } + \frac{13}{5 + 2 \sqrt{3} }$
пжпжпж ответ срочно!! ​

Answer 1

$\dfrac{6}{4-3\sqrt{2} } +\dfrac{13}{5+2\sqrt{3} }=\dfrac{6\cdot(4+3\sqrt{2} )}{(4-3\sqrt{2})(4+3\sqrt{2} ) } +\dfrac{13\cdot(5-2\sqrt{3}) }{(5+2\sqrt{3})(5-2\sqrt{3}) }= \\\\\\=\dfrac{24+18\sqrt{2} }{(4)^{2} -(3\sqrt{2})^{2} } +\dfrac{65-26\sqrt{3} }{(5)^{2} -(2\sqrt{3})^{2}}=\dfrac{24+18\sqrt{2} }{16 -18 } +\dfrac{65-26\sqrt{3} }{25 -12}=\\\\\\=\dfrac{24+18\sqrt{2} }{ -2 } +\dfrac{65-26\sqrt{3} }{13}=\dfrac{13(5-2\sqrt{3}) }{13}-\dfrac{2(12+9\sqrt{2}) }{2}=$

$=5-2\sqrt{3}-12-9\sqrt{2}=-7-2\sqrt{3}-9\sqrt{2}$