Ok, by divide, I presume you are talking about rationalising the denominator.
For info on how to do this, check this site out: http://www.purplemath.com/modules/radicals5.htm
Ok, on to solving:
In the first one, notice how the problem can be rewritten as: #(2*sqrt(3))/(2*sqrt(5))#. This can be rewritten as #(2/2)*(sqrt(3)/(sqrt(5)))#. Since #2/2# is just 1, we can rewrite the problem as
#(1)*(sqrt(3)/(sqrt(5)))#, which is just #(sqrt(3)/(sqrt(5)))#.
From here, we rationalise the denominator by multiplying the expression by #sqrt(5)/sqrt(5)#. We can do this because #sqrt(5)/sqrt(5)# is simply 1, and multiplying something by 1 doesn't change the nature of the expression.
So our expression becomes: #sqrt(3)/(sqrt(5))*sqrt(5)/sqrt(5)#, which simplifies to become #sqrt(15)/5#. Since #sqrt(15)# is not something we can simplify, our final answer remains #sqrt(15)/5#.
Now for the second problem, the procedure is basically the same.
We multiply the expression by #sqrt(10)/sqrt(10)#, so we get: #sqrt(5)/sqrt(10)*sqrt(10)/sqrt(10)#, which simplifies to become #sqrt(50)/sqrt(100)#.
Since #sqrt(100)# simplifies to 10, the expression can be simplified to read: #sqrt(50)/10#.
Now unlike the last problem, this numerator can be simplified, as it is a multiple of 25. #sqrt(50)=5sqrt(2)#
So our expression reads #(5sqrt(2))/10#, which can be simplified to #sqrt(2)/2# for our final answer.
