How do you simplify # [(sqrt2) + (sqrt5)] / [(sqrt2) – (sqrt5)]#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Konstantinos Michailidis Sep 11, 2015 It is #(7+2sqrt2sqrt5)/(-3)# Explanation: #(sqrt2+sqrt5)/(sqrt2-sqrt5)=((sqrt2+sqrt5)*(sqrt2+sqrt5))/((sqrt2-sqrt5)*(sqrt2+sqrt5))=(sqrt2+sqrt5)^2/(2-5)=(2+5+2sqrt2sqrt5)/(-3)=(7+2sqrt2sqrt5)/(-3)# 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 1745 views around the world You can reuse this answer Creative Commons License