If 1-y1-2x-4x2-8x3-16x4-32x5 - 1-y6- where y 1- x 12- then the value of y x is

(1 y)(1+2x+4x^2+8x^3+16x^4+32x^5) = 1 y^6 ⇒ (1 y)(1((2x)^6 1)2x 1)=1 y^6 ⇒ nbsp;(2x)^6 12x 1 = y^6 1y 1 ⇒ y = 2x ∴yx = 2 ...

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If 1-y1+2x+4x...

Question

If $(1 - y) (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + 32 x^{5}) = 1 - y^{6},$ where $y \neq 1, x \neq \frac{1}{2}$ , then the value of $\frac{y}{x}$ is

\frac{1}{2}

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\frac{25}{24}

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\frac{24}{25}

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Solution

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Similar questions

(1 - y) (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + 32 x^{5}) = 1 - y^{6},

y \neq 1, x \neq \frac{1}{2}

\frac{y}{x}

S = \frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}}

T = \frac{8 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})}

x, y, z \neq \pm 1

x + y + z = x y z

S - T

x, y, z > 0

x + y + z = 1,

\frac{2 x}{1 - x} + \frac{2 y}{1 - y} + \frac{2 z}{1 - z}

y (x)

\frac{d y}{d x} + (\frac{2 x + 1}{x}) y = e^{- 2 x}, x > 0

y (1) = \frac{1}{2} e^{- 2}

\frac{37}{13} = 2 + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}

x, y, z

x + y + z

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