Which one of the following is the greatest ____. - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

given,

\sqrt{15} - \sqrt{13}, \sqrt{13} - \sqrt{11}, \sqrt{11} - \sqrt{9} and \sqrt{9} - \sqrt{7}

we get

\sqrt{15} - \sqrt{13}

\frac{\sqrt{15} - \sqrt{13} \times \sqrt{15} + \sqrt{13}}{\sqrt{15} + \sqrt{13}} = \frac{2}{\sqrt{15} + \sqrt{13}}

\sqrt{13} - \sqrt{11}

\frac{\sqrt{13} - \sqrt{11} \times \sqrt{13} + \sqrt{11}}{\sqrt{13} + \sqrt{11}} = \frac{2}{\sqrt{13} + \sqrt{11}}

\sqrt{11} - \sqrt{9}

\frac{\sqrt{11} - \sqrt{9} \times \sqrt{11} + \sqrt{9}}{\sqrt{11} + \sqrt{9}} = \frac{2}{\sqrt{11} + \sqrt{9}}

\sqrt{9} - \sqrt{7}

\frac{\sqrt{9} - \sqrt{7} \times \sqrt{9} + \sqrt{7}}{\sqrt{9} + \sqrt{7}} = \frac{2}{\sqrt{9} + \sqrt{7}}

now, the denominator of

\sqrt{9} + \sqrt{7}

,
Hence, option D is the correct option.

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