Question
Rationales the denominator and simplify: $\frac{2\sqrt6-\sqrt5}{3\sqrt5-2\sqrt6}$
✓
Answer
$\frac{2\sqrt6-\sqrt5}{3\sqrt5-2\sqrt6}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $3\sqrt5+2\sqrt6$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{\big(3\sqrt5-2\sqrt6\big)\big(3\sqrt5+2\sqrt6\big)}$
As we know, $(\text{a}-\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{45-24}$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{21}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{21}$
$=\frac{4\sqrt{30}+9}{21}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $3\sqrt5+2\sqrt6$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{\big(3\sqrt5-2\sqrt6\big)\big(3\sqrt5+2\sqrt6\big)}$
As we know, $(\text{a}-\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{45-24}$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{21}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{21}$
$=\frac{4\sqrt{30}+9}{21}$
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