# How do you simplify the fractional expressions #1/(x-3) - 1/(x+3) - 1/(x-1) +1/(x+1)#?

we just take L.C.M and use the formula (a+b) (a-b)=(a²-b²) after that the brackets are opened and it is simplified by cancelling like terms with opposite signs.

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To simplify the given fractional expression, we can combine the fractions by finding a common denominator. The common denominator for the fractions (x-3), (x+3), (x-1), and (x+1) is (x-3)(x+3)(x-1)(x+1).

Multiplying each fraction by the appropriate factors of the common denominator, we get:

[(x+3)(x-1)(x+1) - (x-3)(x+1)(x-1) - (x-3)(x+3)(x+1) + (x-3)(x+3)(x-1)] / [(x-3)(x+3)(x-1)(x+1)]

Expanding and simplifying the numerator, we have:

[(x^2 + 4x + 3)(x+1) - (x^2 - 4x + 3)(x-1) - (x^2 - 9)(x+1) + (x^2 - 9)(x-1)] / [(x-3)(x+3)(x-1)(x+1)]

Further simplifying the numerator, we get:

[(x^3 + 5x^2 + 7x + 3) - (x^3 - 3x^2 + 3) - (x^3 + x^2 - 9x - 9) + (x^3 - x^2 - 9x + 9)] / [(x-3)(x+3)(x-1)(x+1)]

Combining like terms in the numerator, we have:

[5x^2 + 7x + 3 - 3x^2 + 3 - x^2 + x^2 - 9x - 9 + x^2 - 9x + 9] / [(x-3)(x+3)(x-1)(x+1)]

Simplifying further, we get:

[x^2 - 10x + 3] / [(x-3)(x+3)(x-1)(x+1)]

Therefore, the simplified form of the given fractional expression is (x^2 - 10x + 3) / [(x-3)(x+3)(x-1)(x+1)].

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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