मराठी

If X = √ 5 + √ 3 √ 5 − √ 3 and Y = √ 5 − √ 3 √ 5 + √ 3 Then X + Y +Xy= - Mathematics

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प्रश्न

If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=

पर्याय

  • 9

  • 5

  • 17

  • 7

MCQ

उत्तर

Given that  `x=(sqrt5 +sqrt3)/(sqrt5 - sqrt3)`and `y = (sqrt5 - sqrt3)/(sqrt5 +sqrt3)`.

We are asked to find  `x+y + xy`

Now we will rationalize x. We know that rationalization factor for  `sqrt5 -sqrt3` is `sqrt5 +sqrt3`. We will multiply numerator and denominator of the given expression   `x=(sqrt5 +sqrt3)/(sqrt5 - sqrt3)` by `sqrt5 + sqrt3`, to get

 `x=(sqrt5 +sqrt3)/(sqrt5 - sqrt3) xx (sqrt5 +sqrt3)/(sqrt5 + sqrt3)`

`= ((sqrt5)^2+(sqrt3)^2+ 2 xx sqrt5 xx sqrt3)/((sqrt5)^2 - (sqrt3)^2)`

`= (5+3+2sqrt15)/(5-3)`

`= 4 + sqrt15`

Similarly, we can rationalize y. We know that rationalization factor for   `sqrt5 +sqrt3`is`sqrt5 - sqrt3`. We will multiply numerator and denominator of the given expression   `(sqrt5 - sqrt3)/(sqrt5+sqrt3)`by,`sqrt5 - sqrt3` to get

x = `(sqrt5 - sqrt3)/(sqrt5+sqrt3) xx (sqrt5 - sqrt3)/(sqrt5-sqrt3)`

` = ((sqrt5)^2 + (sqrt3)^2 - 2 xx sqrt5 xx sqrt3)/((sqrt5)^2 - (sqrt3)) `

`= (5+3-2sqrt15)/(5-3)`

`= (8-2sqrt15)/2`

`= 4-sqrt15`

Therefore,

`x+y+xy = 4 +sqrt15 + 4 -sqrt15 +(4+sqrt15) (4-sqrt15)`

`= 4+4 + 16 - 4sqrt15 + 4 sqrt15 - (sqrt15)^2`

` = 24 - 15 `

 =` 9`

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पाठ 3: Rationalisation - Exercise 3.4 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 9
पाठ 3 Rationalisation
Exercise 3.4 | Q 13 | पृष्ठ १७

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