Question

plZ answer this question with steps​

Answer 1

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Answer 2

Given :

$\mathsf{\bigstar\;\;x = 5 + 2\sqrt{6}}$

$\mathsf{\bigstar\;\;xy = 1}$

$\mathsf{\implies (5 + 2\sqrt{6})(y) = 1}$

$\mathsf{\implies y = \dfrac{1}{5 + 2\sqrt{6}}}$

$\mathsf{Multiplying\;and\;Dividing\;with\;5 - 2\sqrt{6},\;We\;get :}$

$\mathsf{\implies y = \dfrac{5 - 2\sqrt{6}}{(5 + 2\sqrt{6})(5 - 2\sqrt{6})}}}$

★  We know that : (a + b)(a - b) = a² + b²

$\mathsf{\implies y = \dfrac{5 - 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2}}}$

$\mathsf{\implies y = \dfrac{5 - 2\sqrt{6}}{25 - (2)^2(\sqrt{6})^2}}}$

$\mathsf{\implies y = \dfrac{5 - 2\sqrt{6}}{25 - 4(6)}}}$

$\mathsf{\implies y = \dfrac{5 - 2\sqrt{6}}{25 - 24}}}$

$\mathsf{\implies y = 5 - 2\sqrt{6}}$

$\mathsf{Now,\;Consider :\;\dfrac{1}{x^2} + \dfrac{1}{y^2}}$

$\mathsf{It\;can\;be\;written\;as : \bigg(\dfrac{1}{x}\bigg)^2 + \bigg(\dfrac{1}{y}\bigg)^2}$

$\mathsf{As :\;xy = 1 \implies x = \dfrac{1}{y}\;and\;y = \dfrac{1}{x}}$

$\mathsf{\implies \dfrac{1}{x^2} + \dfrac{1}{y^2} = \bigg(\dfrac{1}{x}\bigg)^2 + \bigg(\dfrac{1}{y}\bigg)^2 = y^2 + x^2}$

Substituting the values of x and y in x² + y², We get :

$\mathsf{\implies (5 + 2\sqrt{6})^2 + (5 - 2\sqrt{6})^2}$

★  We know that : (a + b)² = a² + 2ab + b²

★  We know that : (a - b)² = a² - 2ab + b²

$\mathsf{\implies (5)^2 + (2\sqrt{6})^2 + 2(5)(2\sqrt{6}) + (5)^2 + (2\sqrt{6})^2 - 2(5)(2\sqrt{6})}$

$\mathsf{\implies 25 + (2)^2(\sqrt{6})^2 + 25 + (2)^2(\sqrt{6})^2}$

$\mathsf{\implies 50 + (4)(6) + (4)(6)}$

$\mathsf{\implies 50 + 24 + 24}$

$\mathsf{\implies 98}$

$\textbf{Answer : \boxed{\mathsf{\dfrac{1}{x^2} + \dfrac{1}{y^2} = 98}}}$