Question

rationalise the denominator 3√2/√3+√6-4√3/√6+√2+√6/√2+√3

Answer 1

(3√2) /(√3+√6) = [ { (3√2) × (-√3+√6) } / {(√3+√6) × (-√3+√6)} ]                         = (- 3√6 + 3√12)/(3) = √12 - √6 = 2√3 - √6. (4√3) /(√2+√6) = [ { (4√3) × (-√2+√6) } / {(√2+√6) × (-√2+√6)} ]                         = (- 4√6 + 4√18)/(4) = √18 - √6 = 3√2 - √6.(√6) /(√3+√2) = [ { (√6) × (-√2+√3) } / {(√3 + √2) × (-√2+√3)} ]                         = (- √12 + √18)/(1) = -√12 + √18 = 3√2 - 2√3. So, (3√2) /(√3+√6)  + (4√3) /(√2+√6) + (√6) /(√3+√2)       = 2√3 - √6. + 3√2 - √6. + 3√2 - 2√3.      = 6√2 - 2√6  

Answer 2

here's your answer !!

√6 / ( √3-√2 )
= √6 ( √3 + √2 ) / ( √3-√2 )( √3 + √2 )
= ( √6 * √3 ) + ( √6 * √2 ) / ( √3 )² - ( √2 )²
= √18 + √12 / ( 3 -2 )
= 3√2 +2√3