Question

answer the questions​

Answer 1

Answer:

we will cross multipy

11×x²+xy+y²

12×x²-xy+y²

Answer 2

$\huge\underline\red\bigstar{\boxed{\boxed{\mathfrak\red{Hello}}}}$

$x = \frac{ \sqrt{7 } + \sqrt{3} }{ \sqrt{7} - \sqrt{3} } \: and \: xy = 1$

. :

$y = \frac{1}{x} = \frac{ \sqrt{7} - \sqrt{3} }{ \sqrt{7} + \sqrt{3} } \\ \\ Now, \\ x + y = \frac{ \sqrt{7} + \sqrt{3} }{ \sqrt{7} - \sqrt{3} } + \frac{ \sqrt{7} - \sqrt{3} }{ \sqrt{7} + \sqrt{3} }$

$= \frac{ (\sqrt{7} + { \sqrt{3} })^{2} + ( \sqrt{7} - { \sqrt{3} })^{2} }{( \sqrt{7} - \sqrt{3})( \sqrt{7} + \sqrt{3} } \\ = \frac{(7 + 3 + 2 \sqrt{7 } \times \sqrt{3} ) + (7 + 3 - 2 \times \sqrt{7} \times \sqrt{3} )}{ { (\sqrt{7} })^{2} - {( \sqrt{3} })^{2} } \\ = \frac{20}{7 - 3} \\ = \frac{20}{4} \\ = 5$

$Now, \\ \frac{ {x}^{2} + xy + {y}^{2} }{ {x}^{2} - xy + {y}^{2} } \\ = \frac{(x + {y})^{2} - xy }{(x + {y})^{2} - 3xy } \\ = \frac{ ({5})^{2} - 1}{( {5})^{2} - 3 \times 1} \\ = \frac{25 - 1}{25 - 3} \\ = \frac{24}{22} \\ = \frac{12}{11} \: [Proved]$

$\huge\mathfrak\red{Hope \: it \: helps \: uhh}$