Question
Rationales the denominator and simplify:$\frac{2\sqrt3-\sqrt5}{2\sqrt{2}+3\sqrt{3}}$
✓
Answer
$\frac{2\sqrt3-\sqrt5}{2\sqrt{2}+3\sqrt{3}}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $2\sqrt{2}-3\sqrt{3}$
$=\frac{\big(2\sqrt3+\sqrt5\big)\big(2\sqrt{2}-3\sqrt{3}\big)}{\big(2\sqrt{2}+3\sqrt{3}\big)\big(2\sqrt{2}-3\sqrt{3}\big)}$
$=\frac{\big(2\sqrt3+\sqrt5\big)\big(2\sqrt{2}-3\sqrt{3}\big)}{8-27}$
$=\frac{\big(4\sqrt6-2\sqrt{10}\big)-18+3\sqrt{15}}{30}$
$=\frac{18-4\sqrt6+2\sqrt{10}-3\sqrt{15}}{19}$
$=\frac{\big(2\sqrt3+\sqrt5\big)\big(2\sqrt{2}-3\sqrt{3}\big)}{\big(2\sqrt{2}+3\sqrt{3}\big)\big(2\sqrt{2}-3\sqrt{3}\big)}$
$=\frac{\big(2\sqrt3+\sqrt5\big)\big(2\sqrt{2}-3\sqrt{3}\big)}{8-27}$
$=\frac{\big(4\sqrt6-2\sqrt{10}\big)-18+3\sqrt{15}}{30}$
$=\frac{18-4\sqrt6+2\sqrt{10}-3\sqrt{15}}{19}$
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