How do you simplify #1/(sqrt2+sqrt3)#?

Question

7380 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-27T05:19:56

Multiply the denominator with #sqrt3-sqrt2# to get

#(sqrt3-sqrt2)/[(sqrt3+sqrt2)*(sqrt3-sqrt2)]= (sqrt3-sqrt2)/((sqrt3)^2-(sqrt2)^2)= (sqrt3-sqrt2)/(1)=sqrt3-sqrt2#

Answer 2 · 2018-03-01T23:39:13

#(1)/(sqrt2+sqrt3)=color(blue)(sqrt3-sqrt2#

Simplify:

#(1)/(sqrt2+sqrt3)#

Rationalize the denominator.

#(1)/(sqrt2+sqrt3)xx(sqrt2-sqrt3)/(sqrt2-sqrt3)#

Simplify.

#(sqrt2-sqrt3)/(sqrt(2)^2-sqrt(3)^2)#

Apply rule: #sqrt(x)^2=x#.

#(sqrt2-sqrt3)/(2-3)#

Simplify.

#(sqrt2-sqrt3)/(-1)#

Simplify.

#-(sqrt2-sqrt3)#

Simplify parentheses.

#-sqrt2+sqrt3=#

#sqrt3-sqrt2#