How do you simplify the expression #sqrt(1/5)-sqrt5#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Cri007 Jul 29, 2017 See below. Explanation: #sqrt(1/5)-sqrt5=1/sqrt(5)-sqrt5=1/sqrt(5)\cdot\sqrt(5)/sqrt(5)-sqrt5=\sqrt(5)/5-sqrt(5)=(sqrt(5)-5sqrt(5))/5=(-4sqrt(5))/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 3108 views around the world You can reuse this answer Creative Commons License