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प्रश्न
Rationalise the denominator of the following:
`1/(sqrt5+sqrt2)`
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उत्तर
The given number is `1/(sqrt5 + sqrt2)`
On rationalising the denominator,
⇒ `1/(sqrt5 + sqrt2) = 1/(sqrt5 + sqrt2) xx (sqrt5 - sqrt2)/(sqrt5 - sqrt2)`
We know that (a + b) (a - b) = a2 - b2
⇒ `1/(sqrt5 + sqrt2) = (sqrt5 - sqrt2)/((sqrt5)^2 - (sqrt2)^2)`
⇒ `1/(sqrt5 + sqrt2) = (sqrt5 - sqrt2)/(5 - 2)`
∴ `1/(sqrt5 + sqrt2) = (sqrt5 - sqrt2)/3`
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संबंधित प्रश्न
Find the value to three places of decimals of the following. It is given that
`sqrt2 = 1.414`, `sqrt3 = 1.732`, `sqrt5 = 2.236` and `sqrt10 = 3.162`
`2/sqrt3`
If x= \[\sqrt{2} - 1\], then write the value of \[\frac{1}{x} . \]
If \[a = \sqrt{2} + 1\],then find the value of \[a - \frac{1}{a}\].
The rationalisation factor of \[2 + \sqrt{3}\] is
Rationalise the denominator of the following:
`(3 + sqrt(2))/(4sqrt(2))`
Rationalise the denominator of the following:
`(sqrt(3) + sqrt(2))/(sqrt(3) - sqrt(2))`
Rationalise the denominator in the following and hence evaluate by taking `sqrt(2) = 1.414, sqrt(3) = 1.732` and `sqrt(5) = 2.236`, upto three places of decimal.
`1/(sqrt(3) + sqrt(2))`
Simplify:
`(8^(1/3) xx 16^(1/3))/(32^(-1/3))`
Simplify:
`(7sqrt(3))/(sqrt(10) + sqrt(3)) - (2sqrt(5))/(sqrt(6) + sqrt(5)) - (3sqrt(2))/(sqrt(15) + 3sqrt(2))`
If `sqrt(2) = 1.414, sqrt(3) = 1.732`, then find the value of `4/(3sqrt(3) - 2sqrt(2)) + 3/(3sqrt(3) + 2sqrt(2))`.
