Question

√2+√3÷3√2-2√3=a-b√6
Factorise​

Answer 1

Question:

$= > \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = a - b \sqrt{6}$

Answer:

$= > \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = a - b \sqrt{6}$

Rationalising L.H.S.,

$= > \frac{( \sqrt{2} + \sqrt{3} )(3 \sqrt{2} + 2 \sqrt{3} ) }{18 - 12} = a - b \sqrt{6}$

$= > \frac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 }{6} = a - b \sqrt{6}$

$= > 12 + 5 \sqrt{6} = 6a - 6b \sqrt{6}$

On comparing both sides,

$= > 6a = 12$

$= > a = 2$

$= > - 6b = 5$

$= > b = \frac{ - 5}{6}$

Hope this answer helps you ^_^ !