Advertisements
Advertisements
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=
विकल्प
9
5
17
7
Advertisements
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`
APPEARS IN
संबंधित प्रश्न
Find:-
`64^(1/2)`
Solve the following equations for x:
`2^(2x)-2^(x+3)+2^4=0`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
If \[8^{x + 1}\] = 64 , what is the value of \[3^{2x + 1}\] ?
`(2/3)^x (3/2)^(2x)=81/16 `then x =
If g = `t^(2/3) + 4t^(-1/2)`, what is the value of g when t = 64?
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
