Solve \(\frac{x 2}{2x-8x^{3}}\)

Question

Simplify the expression

\frac{1}{1 - 4 x ^{2}}

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x = 0, x = \frac{1}{2}, x = - \frac{1}{2}

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

Evaluate

\frac{x \times 2}{2 x - 8 x ^{3}}

Evaluate

\frac{2 x}{2 x - 8 x ^{3}}

Factor the expression

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

For each factor in the denominator,write a new fraction

\frac{?}{2} + \frac{?}{x} + \frac{?}{1 - 2 x} + \frac{?}{1 + 2 x}

Write the terms in the numerator

\frac{A}{2} + \frac{B}{x} + \frac{C}{1 - 2 x} + \frac{D}{1 + 2 x}

Set the sum of fractions equal to the original fraction

\frac{2 x}{2 x ( 1 - 2 x ) ( 1 + 2 x )} = \frac{A}{2} + \frac{B}{x} + \frac{C}{1 - 2 x} + \frac{D}{1 + 2 x}

Multiply both sides

\frac{2 x}{2 x ( 1 - 2 x ) ( 1 + 2 x )} \times 2 x (1 - 2 x) (1 + 2 x) = \frac{A}{2} \times 2 x (1 - 2 x) (1 + 2 x) + \frac{B}{x} \times 2 x (1 - 2 x) (1 + 2 x) + \frac{C}{1 - 2 x} \times 2 x (1 - 2 x) (1 + 2 x) + \frac{D}{1 + 2 x} \times 2 x (1 - 2 x) (1 + 2 x)

Simplify the expression

2 x = (x - 4 x^{3}) A + (2 - 8 x^{2}) B + (2 x + 4 x^{2}) C + (2 x - 4 x^{2}) D

Simplify the expression

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2 x = A x - 4 A x^{3} + 2 B - 8 B x^{2} + 2 C x + 4 C x^{2} + 2 D x - 4 D x^{2}

Group the terms

2 x = - 4 A x^{3} + (- 8 B + 4 C - 4 D) x^{2} + (A + 2 C + 2 D) x + 2 B

Equate the coefficients

⎩ ⎨ ⎧ 0 = - 4 A 0 = - 8 B + 4 C - 4 D 2 = A + 2 C + 2 D 0 = 2 B

Swap the sides

⎩ ⎨ ⎧ - 4 A = 0 - 8 B + 4 C - 4 D = 0 A + 2 C + 2 D = 2 2 B = 0

Solve the equation for A

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⎩ ⎨ ⎧ A = 0 - 8 B + 4 C - 4 D = 0 A + 2 C + 2 D = 2 2 B = 0

Substitute the given value of A into the equation ⎩ ⎨ ⎧ - 8 B + 4 C - 4 D = 0 A + 2 C + 2 D = 2 2 B = 0

⎩ ⎨ ⎧ - 8 B + 4 C - 4 D = 0 0 + 2 C + 2 D = 2 2 B = 0

Removing 0 doesn't change the value,so remove it from the expression

⎩ ⎨ ⎧ - 8 B + 4 C - 4 D = 0 2 C + 2 D = 2 2 B = 0

Solve the equation for B

⎩ ⎨ ⎧ - 8 B + 4 C - 4 D = 0 2 C + 2 D = 2 B = 0

Substitute the given value of B into the equation {- 8 B + 4 C - 4 D = 0 2 C + 2 D = 2

{- 8 \times 0 + 4 C - 4 D = 0 2 C + 2 D = 2

Simplify

{4 C - 4 D = 0 2 C + 2 D = 2

Solve the equation for C

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{C = D 2 C + 2 D = 2

Substitute the given value of C into the equation 2 C + 2 D = 2

2 D + 2 D = 2

Simplify

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4 D = 2

Divide both sides

\frac{4 D}{4} = \frac{2}{4}

Divide the numbers

D = \frac{2}{4}

Cancel out the common factor 2

D = \frac{1}{2}

Substitute the given value of D into the equation C = D

C = \frac{1}{2}

Calculate

⎩ ⎨ ⎧ A = 0 B = 0 C = \frac{1}{2} D = \frac{1}{2}

Solution

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

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x \in \emptyset

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