Question

If $x = 2 + \sqrt{3}, xy = 1, then \frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}} =$

Easy

Solution

$y = \frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} ∴ \frac{x}{\sqrt{2} + \sqrt{x}} \frac{y}{\sqrt{2} - \sqrt{y}} = \frac{x (\sqrt{2} - \sqrt{x})}{2 - x} + \frac{y (\sqrt{2} + \sqrt{y})}{2 - y} \Rightarrow \frac{x (\sqrt{x} - \sqrt{2})}{x - 2} + \frac{y (\sqrt{y} + \sqrt{2})}{2 - y} - \frac{x (\sqrt{x} - \sqrt{2})}{\sqrt{3}} + \frac{y (\sqrt{y} + \sqrt{2})}{\sqrt{3}} = \frac{1}{\sqrt{3}} [x \sqrt{x} + y \sqrt{y} + \sqrt{2} (y - x)]$
$= \frac{1}{\sqrt{3}} [{(2 + \sqrt{3})}^{3 / 2} + {(2 - \sqrt{3})}^{3 / 2} + \sqrt{2} (- 2 \sqrt{3})] = \frac{1}{\sqrt{3}} [\frac{1}{2^{3 / 2}} {(4 + 2 \sqrt{3})}^{3 / 2} + \frac{1}{2^{3 / 2}} {(4 - 2 \sqrt{3})}^{3 / 2} - 2 \sqrt{6}] = \frac{1}{\sqrt{3}} [\frac{1}{2 \sqrt{2}} \{(\sqrt{3} + 1]^{3} + {(\sqrt{3} - 1)}^{3}\} - 2 \sqrt{6}] = \frac{1}{\sqrt{3}} [\frac{1}{2 \sqrt{2}} \{2.3 \sqrt{3} + 6 \sqrt{3}\} - 2 \sqrt{6}] = \frac{1}{\sqrt{3}} (3 \sqrt{6} - 2 \sqrt{6}) = \sqrt{2} .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME