If x-2-x-1 then the value of x2-x-2-x2 1-x is

The correct option is D 1Given x 2x=1 #160; Multiply by x both sides #160; x^2 2=x #160; x^2 x=2 #160;So #160; x^2+x 2x^2(1 x)= ...

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Property 2

If x-2/x=1 ...

Question

If $x - \frac{2}{x} = 1$ then the value of $\frac{x^{2} + x - 2}{x^{2} (1 - x)}$ is

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l i m x \to 0 {(\frac{e^{x} + e^{- x} - 2}{x^{2}})}^{{1 / x}^{2}}

(1 + x + x^{2}) (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + . . .

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

1 + x + x^{2} + . . . x^{p} = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})

p

1 + x + x^{2} + . . . . x^{p} = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})

p

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