Question

Please answer with explanation if you know ​

Answer 1

Solution :

$\implies \dfrac{\Big(\dfrac{1}{\sqrt{9}}\:-\:\dfrac{1}{\sqrt{11}}\Big)}{\Big(\dfrac{1}{\sqrt{9}}\:+\:\dfrac{1}{\sqrt{11}}\Big)} \times \dfrac{10\:+\:\sqrt{99}}{A}\: =\:\dfrac{1}{2}$

$\implies \dfrac{\Big(\dfrac{\sqrt{11}\:-\:\sqrt{9}} {\sqrt{9\times11}}\Big)}{\Big(\dfrac{\sqrt{11}\:+\:\sqrt{9}}{\sqrt{9\times11}}\Big)}\times \dfrac{10\:+\:\sqrt{99}}{A}\: =\: \dfrac{1}{2}$

$\implies \dfrac{\sqrt{11}\:-\: \sqrt{9}}{\sqrt{11}\:+\: \sqrt{9}}\times \dfrac{10\:+\:\sqrt{99}}{A}\: =\: \dfrac{1}{2}$

$\implies \dfrac{\sqrt{11}\:-\: \sqrt{9}}{\sqrt{11}\:+\: \sqrt{9}}\times\dfrac{\sqrt{11}\:-\: \sqrt{9}} {\sqrt{11}\:-\: \sqrt{9}}\times\dfrac{10\:+\:\sqrt{99}}{A}\: =\: \dfrac{1}{2}$

$\implies \dfrac{(\sqrt{11}\:-\:\sqrt{9})^{2}}{(\sqrt{11})^{2}\:-\:(\sqrt{9})^{2}}\times \dfrac{10\:+\:\sqrt{99}}{A}\: =\: \dfrac{1}{2}$

$\implies \dfrac{(11\:+\:9\:-\:2\sqrt{99})}{11\: - \: 9}\times \dfrac{10\:+\:\sqrt{99}}{A}\: =\: \dfrac{1}{2}$

$\implies \dfrac{(20\: - \: 2\sqrt{99})}{2}\times\dfrac{10\:+\:\sqrt{99}}{A}\: =\: \dfrac{1}{2}$

$\implies \dfrac{\cancel{2}(10\: - \:\sqrt{99})}{\cancel{2}}\times\dfrac{10\:+\:\sqrt{99}}{A}\: =\: \dfrac{1}{2}$

$\implies \dfrac{10^{2}\:-\:(\sqrt{99})^{2}}{A}\:=\: \dfrac{1}{2}$

$\implies \dfrac{100\: - \: 99}{A}\: =\: \dfrac{1}{2}$

$\implies \dfrac{1}{A}\: = \:\dfrac{1}{2}$

$\implies A\: = \: 2$

ANSWER

$\boxed{A\: = \: 2}$

Answer 2

Solution !

We have to find the value of A

$\sf \dfrac{\Big[\dfrac{1}{\sqrt{9}}-\dfrac{1}{\sqrt{11}}\bigg]}{\Big[\dfrac{1}{\sqrt{9}}+\dfrac{1}{\sqrt{11}} \Big]}\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\\mapsto\sf \dfrac{\Big[\dfrac{\sqrt{11}-\sqrt{9}}{\sqrt{99}}\Big]}{\Big[\dfrac{\sqrt{11}+\sqrt{9}}{\sqrt{99}}\Big]}\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\\mapsto\sf \Big[\dfrac{ \sqrt{11}-\sqrt{9}}{\cancel{\sqrt{99}}}\Big]\times \bigg[\dfrac{\cancel{\sqrt{99}}}{\sqrt{11}+\sqrt{9}}\bigg]\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\ \mapsto\sf \dfrac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\ \mapsto\sf \dfrac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}\times \bigg[\dfrac{\sqrt{11} - \sqrt{9}}{\sqrt{11} - \sqrt{9}}\bigg]\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\ \mapsto\sf \dfrac{(\sqrt{11} - \sqrt{9})^2}{(\sqrt{11})^2-(\sqrt{9})^2}\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}$

$\sf{\small{\dag\ Formula\ used -}}\\ \\ \sf{\dag (a - b)^2=a^2+b^2 - 2ab}\\ \\ \sf{\dag\ \ a^2-b^2= (a+b)(a-b)}$

$\mapsto\sf \dfrac{(\sqrt{11})^2+(\sqrt{9})^2 - 2\times \sqrt{11}\times \sqrt{9}}{11-9}\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\ \mapsto\sf \dfrac{11+9 - 2\sqrt{99}}{2}\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\ \mapsto\sf \dfrac{20 - 2\sqrt{99}}{2}\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\ \mapsto\sf \dfrac{\cancel{2}(10-\sqrt{99})}{\cancel{2}}\times \dfrac{10+\sqrt{99}}{A}=\dfrac{1}{2}\\ \\ \\ \mapsto\sf (10-\sqrt{99})(10+\sqrt{99})=\dfrac{A}{2}$

$\sf {\dag\ \ a^2-b^2=(a+b)(a-b)}$

$\mapsto\sf (10)^2-(\sqrt{99})^2=\dfrac{A}{2}\\ \\ \\ \mapsto\sf 100-99=\dfrac{A}{2}\\ \\ \\ \mapsto\sf 1= \dfrac{A}{2}\\ \\ \\ \mapsto\sf 2\times 1= A\\ \\ \\ \mapsto\sf A= 2$

$\underline{\boxed{\textsf{\textbf{Value \ of \ A =2}}}}$