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=
पर्याय
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
Prove that:
`(a^-1+b^-1)^-1=(ab)/(a+b)`
Solve the following equation for x:
`2^(x+1)=4^(x-3)`
If `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1),` prove that `a^(q-r)b^(r-p)c^(p-q)=1`
Show that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
If ax = by = cz and b2 = ac, show that `y=(2zx)/(z+x)`
If `27^x=9/3^x,` find x.
If `x=2^(1/3)+2^(2/3),` Show that x3 - 6x = 6
The value of \[\left\{ 2 - 3 (2 - 3 )^3 \right\}^3\] is
If \[2^{- m} \times \frac{1}{2^m} = \frac{1}{4},\] then \[\frac{1}{14}\left\{ ( 4^m )^{1/2} + \left( \frac{1}{5^m} \right)^{- 1} \right\}\] is equal to
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
