If a = sqrt{21} - sqrt{20} and b = sqrt{18} - | vedclass

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(D) We are given $a = \sqrt{21} - \sqrt{20}$ and $b = \sqrt{18} - \sqrt{17}$.To compare $a$ and $b$,we rationalize the expressions:$a = \frac{(\sqrt{2

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$a < b$

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(D) We are given $a = \sqrt{21} - \sqrt{20}$ and $b = \sqrt{18} - \sqrt{17}$.To compare $a$ and $b$,we rationalize the expressions:$a = \frac{(\sqrt{21} - \sqrt{20})(\sqrt{21} + \sqrt{20})}{\sqrt{21} + \sqrt{20}} = \frac{21 - 20}{\sqrt{21} + \sqrt{20}} = \frac{1}{\sqrt{21} + \sqrt{20}}$.Similarly,$b = \frac{(\sqrt{18} - \sqrt{17})(\sqrt{18} + \sqrt{17})}{\sqrt{18} + \sqrt{17}} = \frac{18 - 17}{\sqrt{18} + \sqrt{17}} = \frac{1}{\sqrt{18} + \sqrt{17}}$.Since $\sqrt{21} + \sqrt{20} > \sqrt{18} + \sqrt{17}$,it follows that $\frac{1}{\sqrt{21} + \sqrt{20}} Therefore,$a < b$.

If $a = \sqrt{21} - \sqrt{20}$ and $b = \sqrt{18} - \sqrt{17}$,then

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