Question

please answer this question

answer will be marked as brainliest

Answer 1

the answer is option D (all are equal) .


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Answer 2

(21)

Given : √22 + √21, √23 + √20, √24 + √19.

On rationalizing, we get

$= > \frac{(\sqrt{22} + \sqrt{21})(\sqrt{22} - \sqrt{21})}{(\sqrt{22} - \sqrt{21)}} , \frac{(\sqrt{23} + \sqrt{20})(\sqrt{23} - \sqrt{20})}{(\sqrt{23} - \sqrt{20})}, \frac{(\sqrt{24} + \sqrt{19})(24 - \sqrt{19})}{(\sqrt{24} - \sqrt{19})}$

$= > \frac{1}{\sqrt{22} - \sqrt{21}}, \frac{3}{\sqrt{23} - \sqrt{20}}, \frac{5}{\sqrt{24} - \sqrt{19}}$

Now, compare the denominators.

⇒ (√24 - √19) > (√23 - √20) > (√22 - √21)

⇒ (1/√24 - √19) < (3/√23 - √20) < (5/√22 - √21)

⇒ (√24 + √19) < (√23 + √20) < (√22 + √21).

Therefore, the greatest is √22 + √21.

Hope this helps!