(x+1x) = 2,(x+1x) = 2,\left( {x + \frac{1}{x}} \right){\text{ = 2,}} then (x−1x)" role="presentation">(x−1x)(x−1x)\left( {x - \frac{1}{x}} \right) is equal to = ?" /> (x+1x) = 2,(x+1x) = 2,\left( {x + \frac{1}{x}} \right){\text{ = 2,}} then (x−1x)" role="presentation">(x−1x)(x−1x)\left( {x - \frac{1}{x}} \right) is equal to = ?" /> (x+1x) = 2,(x+1x) = 2,\left( {x + \frac{1}{x}} \right){\text{ = 2,}} then (x−1x)" role="presentation">(x−1x)(x−1x)\left( {x - \frac{1}{x}} \right) is equal to = ?" />

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If $(x + \frac{1}{x}) = 2,$ $\left( {x + \frac{1}{x}} \right){\text{ = 2,}}$ then $(x - \frac{1}{x})$ $\left( {x - \frac{1}{x}} \right)$ is equal to = ?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} (x + \frac{1}{x}) = 2 \\ \Rightarrow {(x + \frac{1}{x})}^{2} = 2^{2} \\ \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 4 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = 2 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} - 2. x . \frac{1}{x} \\ = 2 - 2 = 0 \\ \Rightarrow {(x - \frac{1}{x})}^{2} =  0 \\ \Rightarrow x - \frac{1}{x} = 0 \end{aligned}

$\eqalign{ & \left( {x + \frac{1}{x}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{2}}^2} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 4 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} \cr & \,\,\,\,\,\,\,\, = 2 - 2 = 0 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = 0}} \cr & \Rightarrow x - \frac{1}{x} = 0 \cr}$

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then

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

$\left( {x - \frac{1}{x}} \right)$ is equal to = ?", "text": "If

(x + \frac{1}{x}) = 2,

$\left( {x + \frac{1}{x}} \right){\text{ = 2,}}$ then

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

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

Solution:

\begin{aligned} (x + \frac{1}{x}) = 2 \\ \Rightarrow {(x + \frac{1}{x})}^{2} = 2^{2} \\ \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 4 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = 2 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} - 2. x . \frac{1}{x} \\ = 2 - 2 = 0 \\ \Rightarrow {(x - \frac{1}{x})}^{2} =  0 \\ \Rightarrow x - \frac{1}{x} = 0 \end{aligned}

$\eqalign{ & \left( {x + \frac{1}{x}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{2}}^2} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 4 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} \cr & \,\,\,\,\,\,\,\, = 2 - 2 = 0 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = 0}} \cr & \Rightarrow x - \frac{1}{x} = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} (x + \frac{1}{x}) = 2 \\ \Rightarrow {(x + \frac{1}{x})}^{2} = 2^{2} \\ \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 4 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = 2 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} - 2. x . \frac{1}{x} \\ = 2 - 2 = 0 \\ \Rightarrow {(x - \frac{1}{x})}^{2} =  0 \\ \Rightarrow x - \frac{1}{x} = 0 \end{aligned}

$\eqalign{ & \left( {x + \frac{1}{x}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{2}}^2} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 4 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} \cr & \,\,\,\,\,\,\,\, = 2 - 2 = 0 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = 0}} \cr & \Rightarrow x - \frac{1}{x} = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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