If x-1- 2-1- 2-1-2-x- x- 2- then the value of x is -

The correct option is D 1 x=12 12 12 x ⇒ x=12 2 x3 2x ⇒ x=3 2x4 3x On solving, we get x=1 So, Option (A) ...

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Substitution Method to Remove Indeterminate Form

If x=1/ 2-1...

Question

If $x = \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - x}}}, (x \neq 2)$ , then the value of x is ________?

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If \x+\frac{1}{x}1/x = 5\), then the value of $\frac{3 x^{2} + 3}{2 x^{2} + 5 x + 2}$ = ?

x^{2} - x + 1 = 0

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}

x^{2} - x + 1 = 0

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}

x^{2} - x + 1 = 0

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}

\frac{2 + \frac{1}{3 \frac{4}{5}}}{2 + \frac{1}{3 + \frac{1}{1 + \frac{1}{4}}}} = x

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