Помогите решить 535,536 пожалуйста

Question 1

Question 2

535.
$\frac{a}{3 \sqrt{5} } = \frac{a \sqrt{5} }{3 \sqrt{5} \sqrt{5} } = \frac{a \sqrt{5} }{3( \sqrt{5})^2} = \frac{a \sqrt{5} }{3*5} = \frac{a \sqrt{5} }{15} \\ \\ \frac{4b}{5 \sqrt{c} } = \frac{4b* \sqrt{c} }{5*( \sqrt{c} )^2} = \frac{4b \sqrt{c} }{5c} \\ \\ \frac{6}{ \sqrt{a+b}} = \frac{6 \sqrt{a+b} }{( \sqrt{a+b} )^2} = \frac{6 \sqrt{a+b} }{a+b} \\ \\ \frac{4}{ \sqrt{x-y} } = \frac{4 \sqrt{x-y} }{x-y} \\ \\$
$\frac{16}{ \sqrt{a} + \sqrt{6} } = \frac{16( \sqrt{a}- \sqrt{6} ) }{( \sqrt{a}+ \sqrt{6})( \sqrt{a} - \sqrt{6}) } = \frac{16 \sqrt{a}- 16\sqrt{6} }{( \sqrt{a} )^2- ( \sqrt{6})^2 } = \frac{16 \sqrt{a} -16 \sqrt{6} }{a-6} \\ \\ \frac{a}{ \sqrt{c}-4} = \frac{a (\sqrt{c}+4) }{( \sqrt{c}-4)( \sqrt{c} +4) } = \frac{a \sqrt{c}+4a }{c-16}$

536.
$\frac{18}{2 \sqrt{3}+3 } = \frac{18(2 \sqrt{3}-3) }{(2 \sqrt{3} +3)( 2 \sqrt{3}-3) } = \frac{36 \sqrt{3}-54 }{(2 \sqrt{3})^2- 9} = \frac{3(12 \sqrt{3}-18) }{3} = 12 \sqrt{3} -18$

$\frac{11}{3 \sqrt{5} -1} = \frac{11(3 \sqrt{5}+1)}{(3 \sqrt{5}-1)(3 \sqrt{5} +1)} = \frac{33 \sqrt{5} +11}{3^2 *5-1^2} = \frac{33 \sqrt{5}+11 }{44} = \frac{11(3 \sqrt{5}+1) }{11*4} = \\ \\ = \frac{3 \sqrt{5}+1 }{4}$

$\frac{16}{ \sqrt{2}+ \sqrt{3} +1 } = \frac{16( \sqrt{2}- \sqrt{3} -1) }{( \sqrt{2}+ \sqrt{3}+1)( \sqrt{2} - \sqrt{3} -1) } = \\ \\ = \frac{16 \sqrt{2} -16 \sqrt{3}-16 }{2- \sqrt{6}- \sqrt{2}+ \sqrt{6} -3- \sqrt{3} + \sqrt{2}- \sqrt{3}-1 } = \frac{-2(8 \sqrt{2}+8 \sqrt{3}+8) }{-2} = \\ \\ =8( \sqrt{2} + \sqrt{3} +1)$