Rationalizing!?

(2√7)/(√3-√5)

The trick here is just to multiply by the conjugate of the denominator in both the numerator and denominator....so we have.....

[(2√7)/(√3-√5)] *[(√3+√5) / (√3+√5)] =

[2√(21) + 2√(35)] / (3-5) =

[(2)[√(21) + √(35)]] / (-2)  =

-[√(21) + √(35)]

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The next one is similar........

(4x)/(3-√6) * [ (3+√6) / (3+√6)] =

[(4x)(3+√6)]/[9 -6] =

[12x + 4x√6] / [3]

2sqrt7 / (sqrt3 - sqrt5)

(2sqrt7) / (sqrt3 - sqrt5) * [(sqrt3 + sqrt5) / (sqrt3 + sqrt5)] -->

2(sqrt(21) + sqrt(35)) / (3 - 5) -->

2(sqrt(21) + sqrt(35)) / (-2) -->

-(sqrt(21) + sqrt(35)).

4x / (3 - sqrt6)

4x / (3 - sqrt6) * [(3 + sqrt6) / (3 + sqrt6)] -->

4x(3 + sqrt6) / (9 - 6) -->

12x + 4sqrt6 / (3) -->

4x + (4/3)sqrt6.

okay so apparently Aziz was right...

Sorry, Nataszaa, but I believe my answer to the second one was correct......we could "simplify" it further as:

4x + (4/3)x√(6)    (I actually prefer it as a single fraction, but it's arbitrary)

(I believe Aziz forgot to distribute the x over the second term)