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प्रश्न
Write the reciprocal of \[5 + \sqrt{2}\].
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उत्तर
Given that,`5+sqrt2` it’s reciprocal is given as
`1/(5+sqrt2)`
It can be simplified by rationalizing the denominator. The rationalizing factor of `5+sqrt2` is ` 5 - sqrt2`, we will multiply numerator and denominator of the given expression `1/(5+sqrt2)`by, `5-sqrt2` to get
`1/(5+sqrt2) xx (5-sqrt2)/(5-sqrt2) = (5-sqrt2)/((5)^2 - (sqrt2)^2)`
`= (5-sqrt2) /( 25-2)`
` = (5- sqrt2 ) / 23 `
Hence reciprocal of the given expression is `(5- sqrt2 ) / 23 `.
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संबंधित प्रश्न
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = `c/d`. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
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Simplify \[\sqrt{3 - 2\sqrt{2}}\].
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