Advertisements
Advertisements
प्रश्न
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
विकल्प
\[2\sqrt{6}\]
\[2\sqrt{5}\]
24
20
Advertisements
उत्तर
Given that `x = sqrt6 +sqrt5 ` .Hence `1/x`is given as
`1/x = 1/(sqrt6+sqrt5)`.We need to find `x^2 +1/x^2 - 2`
We know that rationalization factor for `sqrt6 +sqrt5` is`sqrt6 -sqrt5`. We will multiply numerator and denominator of the given expression `1/(sqrt6+sqrt5)`by `sqrt6 -sqrt5`, to get
`1/x = 1/(sqrt6+sqrt5) xx (sqrt6-sqrt5)/(sqrt6-sqrt5) `
` = (sqrt6-sqrt5)/((sqrt6)^2 - (sqrt5)^2)`
` = (sqrt6 - sqrt5)/(6-5)`
` = sqrt6 - sqrt5.`
We know that `(x-1/x)^2 = x^2 + 1/x^2 - 2 ` therefore,
`x^2 + 1/x^2 - 2 = (x-1/x)^2 `
` = (sqrt 6 + sqrt5 - (sqrt6 - sqrt5))^2`
` = (sqrt6 + sqrt5 - sqrt6 +sqrt5)^2`
` = (2sqrt5)^2`
`= 20`
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^-4/y^-10)^(5/4)`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
The value of m for which \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\] is
If g = `t^(2/3) + 4t^(-1/2)`, what is the value of g when t = 64?
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
The simplest rationalising factor of \[\sqrt[3]{500}\] is
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=
The positive square root of \[7 + \sqrt{48}\] is
