How do I simplify this?

Split the radicals of fractions into a fraction of radicals:

#color(white)=sqrt(2/3)+sqrt(3/2)+2sqrt(1/24)#

#=sqrt2/sqrt3+sqrt3/sqrt2+2*sqrt1/sqrt24#

Simplify a little:

#=sqrt2/sqrt3+sqrt3/sqrt2+2*1/sqrt24#

#=sqrt2/sqrt3+sqrt3/sqrt2+2/sqrt24#

#=sqrt2/sqrt3+sqrt3/sqrt2+2/(2sqrt6)#

#=sqrt2/sqrt3+sqrt3/sqrt2+1/sqrt6#

Rationalize the denominator or each fraction:

#=sqrt2/sqrt3+sqrt3/sqrt2+1/sqrt6color(red)(*sqrt6/sqrt6)#

#=sqrt2/sqrt3+sqrt3/sqrt2+sqrt6/6#

#=sqrt2/sqrt3+sqrt3/sqrt2color(red)(*sqrt2/sqrt2)+sqrt6/6#

#=sqrt2/sqrt3+sqrt6/2+sqrt6/6#

#=sqrt2/sqrt3color(red)(*sqrt3/sqrt3)+sqrt6/2+sqrt6/6#

#=sqrt6/3+sqrt6/2+sqrt6/6#

Get a common denominator and simplify:

#=(2sqrt6)/6+(3sqrt6)/6+sqrt6/6#

#=(2sqrt6+3sqrt6+sqrt6)/6#

#=(6sqrt6)/6#

#=sqrt6#

That's the simplification. You can check this using a calculator:

Hope this helped!

Simplify:

#sqrt(2/3)+sqrt(3/2)+2sqrt(1/24)#

Apply rule #sqrt(a/b)=sqrta/sqrtb#.

#sqrt2/sqrt3+sqrt3/sqrt2+(2sqrt1)/sqrt24#

Rationalize the denominator in the first term.

#(sqrt2sqrt3)/(sqrt3sqrt3)+sqrt3/sqrt2+(2sqrt1)/sqrt24#

Apply rule #sqrtasqrta=a#.

#(sqrt2sqrt3)/3+sqrt3/sqrt2+(2sqrt1)/sqrt24#

Apply rule #sqrtasqrtb=sqrt(ab)#.

#sqrt6/3+sqrt3/sqrt2+(2sqrt1)/sqrt24#

Rationalize the denominator in the second term.

#sqrt6/3+(sqrt3sqrt2)/(sqrt2sqrt2)+(2sqrt1)/sqrt24#

Apply rule #sqrtasqrta=a#.

#sqrt6/3+(sqrt3sqrt2)/2+(2sqrt1)/sqrt24#

Apply rule #sqrtasqrtb=sqrtab#.

#sqrt6/3+sqrt6/2+(2sqrt1)/sqrt24#

Rationalize the denominator in the last term.

#sqrt6/3+sqrt6/2+(2sqrt1sqrt24)/(sqrt24sqrt24)#

Apply rule #sqrtasqrta=a#.

#sqrt6/3+sqrt6/2+(2sqrt1sqrt24)/24#

Apply rule #sqrtasqrtb=sqrt(ab)#.

#sqrt6/3+sqrt6/2+(2sqrt24)/24#

Prime factorize #sqrt24#.

#sqrt6/3+sqrt6/2+(2sqrt(2^2xx2xx3))/24#

Apply rule #sqrt(a^2)=a#.

#sqrt6/3+sqrt6/2+(2xx2sqrt6)/24#

Simplify.

#sqrt6/3+sqrt6/2+(4sqrt6)/24#

Simplify #(4sqrt6)/24# to #sqrt6/6#.

#sqrt6/3+sqrt6/2+sqrt6/6#

The least common denominator (LCD) is #6#.

Multiply the first two terms by a fractional form of #1# so that each has a denominator of #6#.

#sqrt6/3xxcolor(magenta)2/color(magenta)2+sqrt6/2xxcolor(teal)3/color(teal)3+sqrt6/6#

Simplify.

#(2sqrt6)/6+(3sqrt6)/6+sqrt6/6#

Combine numerators.

#((2sqrt6+3sqrt6+sqrt6))/6#

Simplify.

#(6sqrt6)/6#

Simplify.

#sqrt6#