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प्रश्न
The rationalisation factor of \[2 + \sqrt{3}\] is
विकल्प
\[2 - \sqrt{3}\]
\[2 + \sqrt{3}\]
\[\sqrt{2} - 3\]
\[\sqrt{3} - 2\]
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उत्तर
We know that rationalization factor for `a+sqrt b` is `a-sqrtb` . Hence rationalization factor of `2+sqrt3` is `2-sqrt3 `.
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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?
Represent `sqrt9.3` on the number line.
Simplify the following expressions:
`(sqrt5 - sqrt3)^2`
Rationales the denominator and simplify:
`(3 - sqrt2)/(3 + sqrt2)`
Write the rationalisation factor of \[7 - 3\sqrt{5}\].
Simplify \[\sqrt{3 - 2\sqrt{2}}\].
Rationalise the denominator of the following:
`1/(sqrt7-2)`
Find the value of a and b in the following:
`(5 + 2sqrt(3))/(7 + 4sqrt(3)) = a - 6sqrt(3)`
Rationalise the denominator in the following and hence evaluate by taking `sqrt(2) = 1.414, sqrt(3) = 1.732` and `sqrt(5) = 2.236`, upto three places of decimal.
`6/sqrt(6)`
Simplify:
`[((625)^(-1/2))^((-1)/4)]^2`
