Divide the expression x 2-1- x 2-2-2 x -2- x by x -1- x -2 -A- x 2-1- x 2B- x 2-1- x 2C- x -1- x D- x -1- x

The correct option is C (x+1/x)Given: (x^2+1/x^2+2 2x 2/x)/ ( x+1/x 2 ) Consider the numerator, x^2+1/x^2+2 2x 2/x =( x^2+2+1/x^2) 2( x+1/x) [Using identity: ( ...

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Byju's Answer

Standard VIII

Mathematics

Division of an Expression by a Expression

Divide the ex...

Question

Divide the expression
${x^{2} + \frac{1}{x^{2}} + 2 - 2 x - \frac{2}{x}} by {x + \frac{1}{x} - 2} .$

(x^{2} + \frac{1}{x^{2}})

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

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

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(x^{2} - \frac{1}{x^{2}})

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Solution

The correct option is C $(x + \frac{1}{x})$
Given: $\frac{(x^{2} + \frac{1}{x^{2}} + 2 - 2 x - \frac{2}{x})}{(x + \frac{1}{x} - 2)}$
Consider the numerator,
$x^{2} + \frac{1}{x^{2}} + 2 - 2 x - \frac{2}{x} = (x^{2} + 2 + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x})$
$[Using identity: (a + b)^{2} = a^{2} + 2 a b + b^{2}]$

$= (x + \frac{1}{x})^{2} - 2 (x + \frac{1}{x})$

$= (x + \frac{1}{x}) (x + \frac{1}{x} - 2)$

Therefore, $\frac{(x^{2} + \frac{1}{x^{2}} + 2 - 2 x - \frac{2}{x})}{(x + \frac{1}{x} - 2)} = \frac{(x + \frac{1}{x}) (x + \frac{1}{x} - 2)}{(x + \frac{1}{x} - 2)}$
$= (x + \frac{1}{x})$

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Divide the expression {x2+1x2+2−2x−2x} by {x+1x−2}.

The correct option is C (x+1x)Given: (x2+1x2+2−2x−2x)(x+1x−2) Consider the numerator, x2+1x2+2−2x−2x=(x2+2+1x2)−2(x+1x) [Using identity: (a+b)2=a2+2ab+b2] =(x+1x)2−2(x+1x) =(x+1x)(x+1x−2) Therefore, (x2+1x2+2−2x−2x)(x+1x−2)=(x+1x)(x+1x−2)(x+1x−2) =(x+1x)

Divide the expression
${x^{2} + \frac{1}{x^{2}} + 2 - 2 x - \frac{2}{x}} by {x + \frac{1}{x} - 2} .$