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प्रश्न
Simplify by rationalising the denominator in the following.
`(2sqrt(6) - sqrt(5))/(3sqrt(5) - 2sqrt(6)`
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उत्तर
`(2sqrt(6) - sqrt(5))/(3sqrt(5) - 2sqrt(6)`
= `(2sqrt(6) - sqrt(5))/(3sqrt(5) - 2sqrt(6)) xx (3sqrt(5) + 2sqrt(6))/(3sqrt(5) + 2sqrt(6)`
= `(6sqrt(30) + 24 - 15 - 2sqrt(30))/((3sqrt(5))^2 - (2sqrt(6))^2`
= `(6sqrt(30) + 9 - 2sqrt(30))/(45 - 24)`
= `(4sqrt(30) + 9)/(21)`
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संबंधित प्रश्न
Rationalize the denominator.
`(sqrt 5 - sqrt 3)/(sqrt 5 + sqrt 3)`
Rationalize the denominator.
`2/(3 sqrt 7)`
Rationalise the denominators of : `3/[ sqrt5 + sqrt2 ]`
Simplify by rationalising the denominator in the following.
`(3sqrt(2))/sqrt(5)`
Simplify by rationalising the denominator in the following.
`(5)/(sqrt(7) - sqrt(2))`
In the following, find the values of a and b.
`(sqrt(3) - 1)/(sqrt(3) + 1) = "a" + "b"sqrt(3)`
If x = `((sqrt(3) + 1))/((sqrt(3) - 1)` and y = `((sqrt(3) - 1))/((sqrt(3) + 1)`, find the values of
x2 + y2
If x = `(1)/((3 - 2sqrt(2))` and y = `(1)/((3 + 2sqrt(2))`, find the values of
x3 + y3
Simplify:
`(sqrt(x^2 + y^2) - y)/(x - sqrt(x^2 - y^2)) ÷ (sqrt(x^2 - y^2) + x)/(sqrt(x^2 + y^2) + y)`
Using the following figure, show that BD = `sqrtx`.
