Advertisements
Advertisements
प्रश्न
Simplify by rationalising the denominator in the following.
`(2sqrt(3) - sqrt(6))/(2sqrt(3) + sqrt(6)`
Advertisements
उत्तर
`(2sqrt(3) - sqrt(6))/(2sqrt(3) + sqrt(6)`
= `(2sqrt(3) - sqrt(6))/(2sqrt(3) + sqrt(6)) xx (2sqrt(3) - sqrt(6))/(2sqrt(3) - sqrt(6)`
= `((2sqrt(3) - sqrt(6))^2)/((2sqrt(3))^2 - (sqrt(6))^2`
= `(12 + 6 - 4sqrt(18))/(12 - 6)`
= `(18 - 4sqrt(18))/(6)`
= `(9 - 2sqrt(18))/(3)`
= `(9 - 6sqrt(2))/(3)`
= 3 - 2`sqrt(2)`
APPEARS IN
संबंधित प्रश्न
Rationalise the denominators of : `(2sqrt3)/sqrt5`
Simplify by rationalising the denominator in the following.
`(42)/(2sqrt(3) + 3sqrt(2)`
Simplify by rationalising the denominator in the following.
`(sqrt(5) - sqrt(7))/sqrt(3)`
Simplify by rationalising the denominator in the following.
`(3 - sqrt(3))/(2 + sqrt(2)`
Simplify by rationalising the denominator in the following.
`(7sqrt(3) - 5sqrt(2))/(sqrt(48) + sqrt(18)`
In the following, find the values of a and b:
`(5 + 2sqrt(3))/(7 + 4sqrt(3)) = "a" + "b"sqrt(3)`
In the following, find the value of a and b:
`(sqrt(3) - 1)/(sqrt(3) + 1) + (sqrt(3) + 1)/(sqrt(3) - 1) = "a" + "b"sqrt(3)`
If x = `(4 - sqrt(15))`, find the values of
`x + (1)/x`
If x = `(4 - sqrt(15))`, find the values of:
`(x + (1)/x)^2`
Using the following figure, show that BD = `sqrtx`.
