Question

please tell me this question answer ​

Answer 1

$\frac{ \sqrt{11 } - \sqrt{7} }{ \sqrt{11} + \sqrt{7} } = a + b \sqrt{77}$

Rationalise The Denominator.

As we know

Rationalising Factor of (a+b)=(a-b)

√11+√7=√11-√7

Multiply The Numerator and Denominator With √11-√7

$\frac{ (\sqrt{11} - \sqrt{7} )( \sqrt{11} - \sqrt{7}) }{ (\sqrt{11} + \sqrt{7})( \sqrt{11} - \sqrt{7} )} \\ \\ (x + y) {}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\( x + y) (x - y) = {x}^{2} - {y}^{2} \\ \\ \frac{( \sqrt{11} + \sqrt{7}) {}^{2} }{ \sqrt{11} {}^{2} - \sqrt{7} {}^{2} } \\ \\ \frac{ { \sqrt{11} {} }^{2} + \sqrt{7} {}^{2} + 2 \times \sqrt{11} \times \sqrt{7} }{11 - 7} \\ \\ \frac{11 + 7 + 2 \sqrt{77} }{4} = a + b \sqrt{77} \\ \\ \frac{18 + 2 \sqrt{77} }{4} = a + b \sqrt{77} \\ \\ \frac{18}{4} + \frac{2}{4} \times \sqrt{77} = a + b \sqrt{77}$

By Comparing We get

$a = \frac{18}{4} = \frac{9}{2} \\ \\ b = \frac{2}{4} = \frac{1}{2}$

So

$\huge{ \underline{a = \frac{9}{2} }} \\ \\ \huge{ \underline{b = \frac{1}{2} }}$