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√99 = √9*√11 = 3√11
√44 = √4*√11 = 2√11

√99 - √44+ √11 = 3√11 - 2√11 + √11 = √11(3-2+1) = 2√11

Option B.
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V11(V9-V4+1)=2V11

PUSH B
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\(\sqrt{99}-\sqrt{44}+\sqrt{11} = \sqrt{11}(3-2+1) = 2*\sqrt{11}\)
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The answer is B

√99−√44+√11= 3√11 - 2√11 + √11 = √11 * (3-2+1) = 2*√11
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Bunuel
\(\sqrt{99 }- \sqrt{44 }+ \sqrt{11 }=\)


A. \( \sqrt{11}\)

B. \(2 * \sqrt{11}\)

C. \(\sqrt{66}\)

D. 12

E. \(6 * \sqrt{11}\)


Happy New Year Australia/Northern Territory!

We can try taking out the squares first and see what happens:


\(\sqrt{99 }- \sqrt{44 }+ \sqrt{11 }=3\sqrt{11} - 2\sqrt{11} + \sqrt{11} = 2\sqrt{11}\).

Ans: B
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IMO B
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IMO B

√99−√44+√11
= 3√11−2√11+√11
=2√11
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√99−√44+√11= 3√11- 2√11+√11= 4√11-2√11= 2*√11

So, It is B. :)
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IMO B 2root11

taking root11 common from the equation and then solving it will give the required result.
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answer is b) 2 * underroot 11

3* underroot 11 - 2* underoot 11 + underoot 11

gives 2 underroot 11

so answer is b )
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Correct Answer B

99^(1/2) - 44^(1/2) + 11^(1/2)

\sqrt{99}-\sqrt{44}+\sqrt{11}
=3\sqrt{11}-2\sqrt{11}+\sqrt{11}
=2\sqrt{11}
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√99 - √44 + √11
√9*11 - √4*11 + √11
3√11 - 2√11 + √11
2√11
So the answer will be B
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Bunuel
\(\sqrt{99 }- \sqrt{44 }+ \sqrt{11 }=\)


A. \( \sqrt{11}\)

B. \(2 * \sqrt{11}\)

C. \(\sqrt{66}\)

D. 12

E. \(6 * \sqrt{11}\)


Happy New Year Australia/Northern Territory!

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In the expression, take 9, 4 and 1 out from the exponents in the respective terms. This simplifies the expression to:

3(11)^0.5 - 2(11)^0.5 + (11)^0.5
-> 2(11)^0.5

Thus, IMO option B is correct.
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