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During the war, a flu pandemic killed 20 million people, twice a many as

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1 answer:

andrezito [222]3 years ago

8 0

Answer:

Hi, the sentence It's sadly true.

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(x+1/x-1 - x-1/x+1 - 4x/x^2 +1) + 4x/x^4-1

Lunna [17]

Answer:

$\frac{12x}{x^4-1}$

Explanation:

Given

$(\frac{x+1}{x-1} - \frac{x-1}{x+1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

Required

Solve:

Take L.C.M of the first two fractions

$(\frac{(x+1)(x+1)-(x-1)(x-1)}{(x-1)(x+1)} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

$(\frac{(x+1)(x+1)-(x-1)(x-1)}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

$(\frac{x^2+2x+1-(x^2-2x+1)}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

$(\frac{x^2+2x+1-x^2+2x-1}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

Collect Like Terms

$(\frac{x^2-x^2+2x+2x+1-1}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

$(\frac{4x}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

Solve the expression in bracket

$(\frac{4x(x^2+1) - 4x(x^2-1)}{(x^2-1)(x^2+1)} ) + \frac{4x}{x^4 - 1}$

$(\frac{4x(x^2+1) - 4x(x^2-1)}{x^4-1} ) + \frac{4x}{x^4 - 1}$

$(\frac{4x^3+4x - 4x^3+4x)}{x^4-1} ) + \frac{4x}{x^4 - 1}$

$(\frac{4x^3- 4x^3+4x +4x)}{x^4-1} ) + \frac{4x}{x^4 - 1}$

$\frac{8x}{x^4-1} + \frac{4x}{x^4 - 1}$

Take LCM

$\frac{8x+4x}{x^4-1}$

$\frac{12x}{x^4-1}$

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