Question

rationalize the denominator​

Answer 1

To rationalise the denominator, do the following steps,

$2 \sqrt{3 } + 3 \sqrt{2} \div 2 \sqrt{3} - 3 \sqrt{2}$

$2 \sqrt{3 } + 3 \sqrt{2 } \div 2 \sqrt{3} - 3 \sqrt{2} \times 2 \sqrt{3 } + 3 \sqrt{2} \div 2 \sqrt{3} + 3 \sqrt{2}$

$2 \sqrt{3 } + 3 \sqrt{2 } \div {3}^{2} - {(3 \sqrt{2) }}^{2}$

$2 \sqrt{3 } + 3 \sqrt{2 } \div (6 -18)$

$2 \sqrt{3 } + 3 \sqrt{2 } \div ( - 12)$

Answer 2

Answer:

$- 5 - 2 \sqrt{6 }$

Step-by-step explanation:

$\frac{2 \sqrt{3} + 3 \sqrt{2} }{2 \sqrt{3} - 3 \sqrt{2} } \\ lets \: rationalize = ) \\ ) \frac{2 \sqrt{3} + 3 \sqrt{2} }{2 \sqrt{3} - 3 \sqrt{2}} \times \frac{2 \sqrt{3} + 3 \sqrt{2} }{2 \sqrt{3} + 3 \sqrt{2} } \\ \frac{(x + y)(x + y) = {x}^{2} + 2xy + {y}^{2} }{(x + y)(x - y) = {(x)}^{2} - {(y)}^{2} } \\ = ) \frac{ {(2 \sqrt{3} {)}^{2} + (2 \times 2 \sqrt{3} \times 3 \sqrt{2} )}^{2} + {(3 \sqrt{2} )}^{2} }{ {(2 \sqrt{3} )}^{2} - {(3 \sqrt{2} )}^{2} } \\ = ) \frac{(4 \times 3) + 12 \sqrt{6} + (9 \times 2)}{(4 \times 3) - (9 \times 2)} \\ = ) \frac{12 + 12 \sqrt{6} + 18}{12 - 18} \\ = ) \frac{12 + 12 \sqrt{6} + 18}{ - 6} \\ = ) \frac{6(2 + 2 \sqrt{6} + 3)}{ - 6} \\ = ) - 1(2 + 2 \sqrt{6} + 3) \\ = ) - 2 - 2 \sqrt{6 } - 3 \\ = ) - 5 - 2 \sqrt{6}$

Hope it helped