How do you simplify #1/(sqrt2+sqrt7)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria Jun 21, 2016 #1/(sqrt2+sqrt7)=(sqrt7-sqrt2)/5# Explanation: We simplify #1/(sqrt2+sqrt7)# by rationalizing the denominator i.e. multiplying numerator and denominator by conjugate of denominator. As #(sqrt2+sqrt7)# can be written as #(sqrt7+sqrt2)# and its conjugate is (sqrt7-sqrt2)#, hence #1/(sqrt2+sqrt7)=1/(sqrt7+sqrt2)# = #(1xx(sqrt7-sqrt2))/((sqrt7+sqrt2)(sqrt7-sqrt2))# = #(sqrt7-sqrt2)/(7-2)=(sqrt7-sqrt2)/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 1104 views around the world You can reuse this answer Creative Commons License