How do you simplify #(sqrt5+sqrt2)/sqrt10#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria · Tazwar Sikder Sep 7, 2016 #(sqrt5+sqrt2)/sqrt10# = #sqrt2/2+sqrt5/5# Explanation: To simplify #(sqrt5+sqrt2)/sqrt10#, we need to rationalize denominator. As it is #sqrt10#, #(sqrt5+sqrt2)/sqrt10# can be rationalized by multiplying numerator and denominator by #sqrt10#. Hence #(sqrt5+sqrt2)/sqrt10# = #(sqrt10(sqrt5+sqrt2))/(sqrt10)^2# = #(sqrt50+sqrt20)/10# = #(sqrt(2×5×5)+sqrt(2×2×5))/10# = #(5sqrt2+2sqrt5)/10# = #sqrt2/2+sqrt5/5# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 1632 views around the world You can reuse this answer Creative Commons License