Question

Rationalize the denominator

Answer 1

A. 
$\frac{4 \sqrt{2}+3 \sqrt{3} }{2 \sqrt{2}-5 \sqrt{3}} = \frac{4 \sqrt{2}+3 \sqrt{3} }{2 \sqrt{2}-5 \sqrt{3}} \times \frac{2 \sqrt{2}+5 \sqrt{3} }{2 \sqrt{2}+5 \sqrt{3}}\\ \\ = \frac{(4 \sqrt{2}+3 \sqrt{3}) \times (2 \sqrt{2}+5 \sqrt{3}) }{(2 \sqrt{2}-5 \sqrt{3}) \times (2 \sqrt{2}+5 \sqrt{3})}\\ \\ = \frac{16+20 \sqrt{6}+6 \sqrt{6}+45 }{8-75} \\ \\ = \frac{61+26 \sqrt{6}}{-67} \\ \\ = -\frac{61+26 \sqrt{6}}{67}$

B. 
$\frac{ \frac{ \sqrt{3} }{2}+1 }{1- \frac{ \sqrt{3} }{2}} = \frac{ \frac{ \sqrt{3}+2 }{2} }{ \frac{2- \sqrt{3} }{2}} = \frac{2+ \sqrt{3} }{2- \sqrt{3}} \\ \\ =\frac{2+ \sqrt{3} }{2- \sqrt{3}} \times \frac{2+ \sqrt{3} }{2+ \sqrt{3}} \\ \\ = \frac{(2+ \sqrt{3})^2}{(2- \sqrt{3})(2+ \sqrt{3})} \\ \\ = \frac{4+3+4 \sqrt{3} }{4-3} \\ \\ = \frac{7+4 \sqrt{3} }{1}$

Answer 2

$\frac{4\sqrt2+3\sqrt3}{2\sqrt2-5\sqrt3}\\\\=\frac{(4\sqrt2+3\sqrt3)(2\sqrt2+5\sqrt3)}{(2\sqrt2-5\sqrt3)(2\sqrt2+5\sqrt3)}\\\\=\frac{8*2+15*3+26\sqrt6}{2^2*2-5^2*3}\\\\=\frac{-61}{67}-\frac{26\sqrt6}{67}\\\\.$

$\frac{\frac{\sqrt3}{2}+1}{1-\frac{\sqrt{3}}{2}}\\\\.\frac{(\frac{\sqrt3}{2}+1)(1+\frac{\sqrt{3}}{2})} {(1-\frac{\sqrt{3}}{2})(1+\frac{\sqrt{3}}{2})}\\\\.\frac{\frac{3}{2^2}+1+\sqrt3} {1^2-\frac{3}{2^2}}\\\\7+4\sqrt3$