Advertisement Remove all ads

Simplify 2/(Sqrt5 + Sqrt3) + 1/(Sqrt3 + Sqrt2) + 3/(Sqrt5 + Sqrt2) - Mathematics

Simplify

`2/(sqrt5 + sqrt3) + 1/(sqrt3 + sqrt2) + 3/(sqrt5 + sqrt2)`

Advertisement Remove all ads

Solution

We know that rationalization factor for `sqrt5 + sqrt3`, `sqrt3 + sqrt2` and `sqrt5 + sqrt2` are `sqrt5 - sqrt3`, `sqrt3 - sqrt2` and `sqrt5 - sqrt2` respectively. We will multiply numerator and denominator of the given expression `2/(sqrt5 + sqrt3)`, `1/(sqrt3 + sqrt2)` and `3/(sqrt5 + sqrt2)` by `2 - sqrt3`, `sqrt5 + sqrt3` and `2 + sqrt5`respectively to get

`2/(sqrt5 + sqrt3) xx (sqrt5 - sqrt3)/(sqrt5 - sqrt3) + 1/(sqrt3 + sqrt2) xx (sqrt3 - sqrt2)/(sqrt3 - sqrt2) - 3/(sqrt5 + sqrt2) xx (sqrt5  - sqrt2)/(sqrt5 - sqrt2) = (2sqrt5 - 2sqrt3)/(5 - 3) + (sqrt3 - sqrt2)/(3 - 2) - (3sqrt5 - 3sqrt2)/(5 - 2)`

` = (2sqrt5 - 2sqrt3)/2 + (sqrt3 - sqrt2)/1 - (3sqrt5 - 3sqrt2)/3`

`= sqrt5 - sqrt3 + sqrt3 - sqrt2 - sqrt5 + sqrt2`

= 0

Hence the given expression is simplified to 0

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 3 Rationalisation
Exercise 3.2 | Q 5.3 | Page 14
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×