Please solve 37th question and don't give link

Hi, Taking term (D^{r}) one by one, \sqrt{12 - \sqrt{140}} = \sqrt{12 - \sqrt{2 \times 2 \times 5 \times 7}} = \sqrt{12 - 2 \sqrt{5 \times 7}} = \sqrt{5 + 7 - 2 \sqrt{5 \times 7}} = \sqrt{{(\sqrt{5})}^{2} + {(\sqrt{7})}^{2} - 2 \sqrt{5} \times \sqrt{7}} = \sqrt{{(\sqrt{5} + \sqrt{7})}^{2}} = \sqrt{5} + \sqrt{7} Using {(a \pm b)}^{2} = a^{2} + b^{2} \pm 2 ab \sqrt{8 - \sqrt{60}} = \sqrt{8 - \sqrt{2 \times 2 \times 3 \times 5}} = \sqrt{8 - 2 \sqrt{3} \times \sqrt{5}} = \sqrt{3 + 5 - 2 \sqrt{3} \times \sqrt{5}} = \sqrt{{(\sqrt{5})}^{2} + {(\sqrt{3})}^{2} - 2 \sqrt{5} \times \sqrt{3}} = \sqrt{{(\sqrt{5} - \sqrt{3})}^{2}} = (\sqrt{5} - \sqrt{3}) Also \sqrt{10 + \sqrt{84}} = \sqrt{10 + \sqrt{2 \times 2 \times 7 \times 3}} = \sqrt{10 + 2 \sqrt{7} \times \sqrt{3}} = \sqrt{7 + 3 + 2 \sqrt{7} \times \sqrt{3}} = \sqrt{{(\sqrt{7})}^{2} + {(\sqrt{3})}^{2} + 2 \sqrt{7} \times \sqrt{3}} = \sqrt{{(\sqrt{7} + \sqrt{3})}^{2}} = (\sqrt{7} + \sqrt{3}) Pyutting all we get = \frac{1}{\sqrt{7} + \sqrt{5}} - \frac{1}{\sqrt{5} - \sqrt{3}} - \frac{2}{\sqrt{7} + \sqrt{3}} Rationalizing each denominator = \frac{1}{\sqrt{7} + \sqrt{5}} \times \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} - \sqrt{5}} - \frac{1}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} - \frac{2}{\sqrt{7} + \sqrt{3}} \times \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} - \sqrt{3}} = \frac{\sqrt{7} - \sqrt{5}}{{(\sqrt{7})}^{2} - {(\sqrt{5})}^{2}} - \frac{\sqrt{5} + \sqrt{3}}{{(\sqrt{5})}^{2} - {(\sqrt{3})}^{2}} - \frac{2 (\sqrt{7} - \sqrt{3})}{{(\sqrt{7})}^{2} - {(\sqrt{3})}^{2}} [Using a^{2} - b^{2} = (a - b) (a + b)] = \frac{\sqrt{7} - \sqrt{5}}{7 - 5} - \frac{\sqrt{5} + \sqrt{3}}{5 - 3} - \frac{2 (\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{\sqrt{7} - \sqrt{5}}{2} - \frac{\sqrt{5} + \sqrt{3}}{2} - \frac{(\sqrt{7} - \sqrt{3})}{2} = \frac{\sqrt{7} - \sqrt{5} - \sqrt{5} - \sqrt{3} - \sqrt{7} + \sqrt{3}}{2} = 0

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