(x+1x)(x+1x)\left( {x + \frac{1}{x}} \right) : (x−1x)" role="presentation">(x−1x)(x−1x)\left( {x - \frac{1}{x}} \right) = 5 : 3 the value of x is/are ?" /> (x+1x)(x+1x)\left( {x + \frac{1}{x}} \right) : (x−1x)" role="presentation">(x−1x)(x−1x)\left( {x - \frac{1}{x}} \right) = 5 : 3 the value of x is/are ?" /> (x+1x)(x+1x)\left( {x + \frac{1}{x}} \right) : (x−1x)" role="presentation">(x−1x)(x−1x)\left( {x - \frac{1}{x}} \right) = 5 : 3 the value of x is/are ?" />

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If $(x + \frac{1}{x})$ $\left( {x + \frac{1}{x}} \right)$ : $(x - \frac{1}{x})$ $\left( {x - \frac{1}{x}} \right)$ = 5 : 3 the value of x is/are ?

1±1
2±2
3±3`
40

Answer:- 1

Explanation:-

Solution:

\frac{(x + \frac{1}{x})}{(x - \frac{1}{x})} = \frac{5}{3}

$\frac{{\left( {x + \frac{1}{x}} \right)}}{{\left( {x - \frac{1}{x}} \right)}} = \frac{5}{3}$
By Componendo and Dividendo

\begin{aligned} \Rightarrow \frac{x + \frac{1}{x} + x - \frac{1}{x}}{x + \frac{1}{x} - x + \frac{1}{x}} = \frac{5 + 3}{5 - 3} \\ \Rightarrow \frac{2 x}{\frac{2}{x}} = \frac{8}{2} \\ \Rightarrow x^{2} = 4 \\ \Rightarrow x = \pm 2 \end{aligned}

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(x - \frac{1}{x})

$\left( {x - \frac{1}{x}} \right)$ = 5 : 3 the value of x is/are ?", "text": "If

(x + \frac{1}{x})

$\left( {x + \frac{1}{x}} \right)$ :

(x - \frac{1}{x})

$\left( {x - \frac{1}{x}} \right)$ = 5 : 3 the value of x is/are ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\frac{(x + \frac{1}{x})}{(x - \frac{1}{x})} = \frac{5}{3}

$\frac{{\left( {x + \frac{1}{x}} \right)}}{{\left( {x - \frac{1}{x}} \right)}} = \frac{5}{3}$
By Componendo and Dividendo

\begin{aligned} \Rightarrow \frac{x + \frac{1}{x} + x - \frac{1}{x}}{x + \frac{1}{x} - x + \frac{1}{x}} = \frac{5 + 3}{5 - 3} \\ \Rightarrow \frac{2 x}{\frac{2}{x}} = \frac{8}{2} \\ \Rightarrow x^{2} = 4 \\ \Rightarrow x = \pm 2 \end{aligned}

$\eqalign{ & \Rightarrow \frac{{x + \frac{1}{x} + x - \frac{1}{x}}}{{x + \frac{1}{x} - x + \frac{1}{x}}} = \frac{{5 + 3}}{{5 - 3}} \cr & \Rightarrow \frac{{2x}}{{\frac{2}{x}}} = \frac{8}{2} \cr & \Rightarrow {x^2} = 4 \cr & \Rightarrow x = \pm 2 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution: