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प्रश्न
Rationalise the denominators of:
`[sqrt6 - sqrt5]/[sqrt6 + sqrt5]`
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उत्तर
`[sqrt6 - sqrt5]/[sqrt6 + sqrt5] xx [sqrt6 - sqrt5]/[sqrt6 - sqrt5]`
= `[(sqrt6 - sqrt5)^2]/[(sqrt6)^2 - (sqrt5)^2]`
= `[(sqrt6)^2 + (sqrt5)^2 - 2(sqrt6 xx sqrt5)]/[(sqrt6)^2 - (sqrt5)^2]`
= `[6 + 5 - 2sqrt30 ]/[(sqrt6)^2 - (sqrt5)^2]`
= `[11 - 2sqrt30 ]/[6 - 5]`
= `11 - 2sqrt30`
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संबंधित प्रश्न
Rationalise the denominators of : `3/sqrt5`
Rationalise the denominators of : `[ √3 + 1 ]/[ √3 - 1 ]`
Simplify:
`sqrt2/[sqrt6 - sqrt2] - sqrt3/[sqrt6 + sqrt2]`
Simplify by rationalising the denominator in the following.
`(3sqrt(2))/sqrt(5)`
Simplify by rationalising the denominator in the following.
`(42)/(2sqrt(3) + 3sqrt(2)`
Simplify by rationalising the denominator in the following.
`(3sqrt(5) + sqrt(7))/(3sqrt(5) - sqrt(7)`
Simplify by rationalising the denominator in the following.
`(sqrt(12) + sqrt(18))/(sqrt(75) - sqrt(50)`
In the following, find the values of a and b.
`(sqrt(3) - 1)/(sqrt(3) + 1) = "a" + "b"sqrt(3)`
In the following, find the values of a and b:
`(7sqrt(3) - 5sqrt(2))/(4sqrt(3) + 3sqrt(2)) = "a" - "b"sqrt(6)`
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)`
