x2+1x2=2, x2+1x2=2,\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}} then the value of x−1x" role="presentation">x−1xx−1x\frac{{x - 1}}{x} is?" /> x2+1x2=2, x2+1x2=2,\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}} then the value of x−1x" role="presentation">x−1xx−1x\frac{{x - 1}}{x} is?" /> x2+1x2=2, x2+1x2=2,\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}} then the value of x−1x" role="presentation">x−1xx−1x\frac{{x - 1}}{x} is?" />

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If $\frac{x^{2} + 1}{x^{2}} = 2,$ $\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}}$ then the value of $\frac{x - 1}{x}$ $\frac{{x - 1}}{x}$ is?

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{2} + \frac{1}{x^{2}} = 2 \\ \Rightarrow {(x - \frac{1}{x})}^{2} + 2. x . \frac{1}{x} = 2 \\ \Rightarrow x - \frac{1}{x} = 2 - 2 \\ \Rightarrow x - \frac{1}{x} = 0 \end{aligned}

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then the value of

\frac{x - 1}{x}

$\frac{{x - 1}}{x}$ is?", "text": "If

\frac{x^{2} + 1}{x^{2}} = 2,

$\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}}$ then the value of

\frac{x - 1}{x}

$\frac{{x - 1}}{x}$ is?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x^{2} + \frac{1}{x^{2}} = 2 \\ \Rightarrow {(x - \frac{1}{x})}^{2} + 2. x . \frac{1}{x} = 2 \\ \Rightarrow x - \frac{1}{x} = 2 - 2 \\ \Rightarrow x - \frac{1}{x} = 0 \end{aligned}

$\eqalign{ & {x^2} + \frac{{{\text{ }}1}}{{{x^2}}} = 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} + 2.x.\frac{1}{x} = 2 \cr & \Rightarrow x - \frac{1}{x} = 2 - 2 \cr & \Rightarrow x - \frac{1}{x} = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution: