Question

How do you solve `(2sqrt3)/sqrt6`?

Answer 1

`(2sqrt(3))/(sqrt(6))=color(blue)(sqrt(2))`

Explanation:

`(color(lime)2sqrt(3))/(color(magenta)sqrt(6)`

`color(white)("XXX")=(color(lime)(sqrt(2) * sqrt(2)) * sqrt(3))/(color(magenta)(sqrt(2) * sqrt(3)))`

`color(white)("XXX")=sqrt(2)`

Answer 2

`(2sqrt(3))/(sqrt(6))=(2sqrt(3))/(sqrt(3)sqrt(2))=(2)/(sqrt(2))=sqrt2`

Answer 3

`sqrt(2)`

Explanation:

`(2sqrt3)/sqrt6`

`rArr (2sqrt3)/sqrt6 xxsqrt6/sqrt6`

`rArr =(2sqrt(18))/sqrt36`

`rArr (2*sqrt(9)*sqrt(2) )/(6)`

`=> ( 6sqrt(2)) / 6`

`rArr sqrt(2)`.

Answer 4

`=>sqrt2`

Explanation:

Because of the radical law

`sqrtab=sqrta*sqrtb`, we can rewrite this expression as

`(2sqrt3)/(sqrt2*sqrt3)`

Cancelling out common terms

`(2cancel(sqrt3))/(sqrt2*cancel(sqrt3)`

`=>2/sqrt2`

The convention is to not have an irrational number in the denominator, so let's multiply the top and bottom by `sqrt2`.

`(2*sqrt2)/(sqrt2)^2`

`(2sqrt2)/2`

`(cancel2sqrt2)/cancel2`

`=>sqrt2`

Hope this helps!