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प्रश्न
In the following, find the values of a and b:
`(sqrt(2) + sqrt(3))/(3sqrt(2) - 2sqrt(3)) = "a" - "b"sqrt(6)`
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उत्तर
`(sqrt(2) + sqrt(3))/(3sqrt(2) - 2sqrt(3)`
= `(sqrt(2) + sqrt(3))/(3sqrt(2) - 2sqrt(3)) xx (3sqrt(2) + 2sqrt(3))/(3sqrt(2) + 2sqrt(3)`
= `((sqrt(2) + sqrt(3))(3sqrt(2) + 2sqrt(3)))/((3sqrt(2))^2 - (2sqrt(3))^2`
= `(sqrt(2)(3sqrt(2) + 2sqrt(3)) + sqrt(3)(3sqrt(2) + 2sqrt(3)))/((9 xx 2) - (4 xx 3))`
= `((3 xx 2 + 2sqrt(6)) + (3sqrt(6) + 2 xx 3))/(18 - 12)`
= `(6 + 2sqrt(6) + 3sqrt(6) + 6)/(6)`
= `(12 + 5sqrt(6))/(6)`
= `2 - (-5/6)sqrt(6)`
= `"a" - "b"sqrt(6)`
Hence, a = 2 and b = `-(5)/(6)`.
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संबंधित प्रश्न
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Simplify by rationalising the denominator in the following.
`(2sqrt(6) - sqrt(5))/(3sqrt(5) - 2sqrt(6)`
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 values of a and b:
`(7sqrt(3) - 5sqrt(2))/(4sqrt(3) + 3sqrt(2)) = "a" - "b"sqrt(6)`
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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)`
