`sqrt(125) + 5 sqrt45 -2 sqrt(245) + sqrt20 = 8sqrt5`

Prove that (a) 3 sqrt48 - 4 sqrt75 + 5 sqrt(192) = 32 sqrt3 (b) sqrt72 + 3 sqrt50 - sqrt(512) + sqrt32) = 9 sqrt2 (c) 2 sqrt(125) + 5 sqrt45 - 2 sqrt(245) - 6 sqrt20 = - sqrt5

sqrt5 sqrt(-125)=

sqrt(45)-3sqrt(20)+4sqrt(5)

Which value among sqrt(11) + sqrt(5), sqrt(14) + sqrt(2), sqrt(8) + sqrt(8) is the largest?

Do as directed: (i) Add : sqrt(125 + 2 sqrt(27) and - 5 sqrt(5) - sqrt(3) (ii) Add: sqrt(7) - sqrt(11) and sqrt(5) - sqrt(11) + sqrt(13) (iii) Multiply : 2 sqrt(2) by 5 sqrt(2) (iv) Multiply : (-3 + sqrt(5)) by 3 (v) Divide : 7 sqrt(5) by - 14 sqrt(125) Divide : 7 sqrt(5) by -14 sqrt(125) (vi) Divide : 2 sqrt(216) - 3 sqrt(27) by 3

The sum of the first n terms of the series 1/(sqrt(2) + sqrt(5)) + 1/(sqrt(5) + sqrt(8)) + 1/(sqrt(8) + sqrt(11)) +... is

Show that (sqrt5 + sqrt3)/(sqrt5 - sqrt3) - (sqrt5 - sqrt3)/(sqrt5 + sqrt3) = 2 sqrt15

8sqrt(5)+sqrt(20)-sqrt(125)

The value of int_5^(10)(sqrt(x+sqrt(20 x-100))+sqrt(x-sqrt(20 x-100)))dx"i s" (1) 10sqrt(5) (2) 5sqrt(5) (3) 10sqrt(2) (4) 8sqrt(2)