How do you simplify #(1)/(x) - (2)/(x^2 + x) + (3)/(x^3 - x^2)#?

Answer 1
The key to this is to make the denominator the same for all 3 fractions. For this, you have to find the least common multiple of the 3 denominators. First denominator is just #x#. Second denominator is #x^2+x=x(x+1)#. Third denominator is #x^3-x^2=x(x^2-x)=x*x(x-1)#. The least common multiple is #x*x*(x+1)(x-1)=x^4-x^2#.
For first fraction to have the least common multiple as the denominator we multiply both the numerator and the denominator by #x^3-x#. #(1*(x^3-x))/(x*(x^3-x))=(x^3-x)/(x^4-x^2)#. For second fraction we multiply both the numerator and the denominator by #x^2-x#. #(2(x^2-x))/((x^2+x)(x^2-x))=(2x^2-2x)/(x^4-x^2)#. For third fraction we multiply both the numerator and the denominator by #x+1#. #(3(x+1))/((x^3-x^2)(x+1))=(3x+3)/(x^4-x^2)#.
Putting all these together: #(x^3-x)/(x^4-x^2)-(2x^2-2x)/(x^4-x^2)+(3x+3)/(x^4-x^2)=(x^3-2x^2+4x+3)/(x^4-x^2)#
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Answer 2

To simplify the expression (1)/(x) - (2)/(x^2 + x) + (3)/(x^3 - x^2), we need to find a common denominator for all the fractions. The common denominator will be x(x^2 + x)(x - 1).

Multiplying the first fraction by (x^2 + x)(x - 1)/(x^2 + x)(x - 1), the second fraction by x(x - 1)/(x(x - 1)), and the third fraction by x(x^2 + x)/(x(x^2 + x)), we get:

(x^2 + x)(x - 1)/(x(x^2 + x)(x - 1)) - 2x(x - 1)/(x(x^2 + x)(x - 1)) + 3x(x^2 + x)/(x(x^2 + x)(x - 1))

Simplifying the numerators, we have:

(x^3 - x^2 + x^2 - x)/(x(x^2 + x)(x - 1)) - 2x(x - 1)/(x(x^2 + x)(x - 1)) + 3x(x^2 + x)/(x(x^2 + x)(x - 1))

Combining like terms in the numerator, we get:

(x^3 - x)/(x(x^2 + x)(x - 1)) - 2x(x - 1)/(x(x^2 + x)(x - 1)) + 3x(x^2 + x)/(x(x^2 + x)(x - 1))

Now, we can combine the fractions:

(x^3 - x - 2x(x - 1) + 3x(x^2 + x))/(x(x^2 + x)(x - 1))

Simplifying the numerator further:

(x^3 - x - 2x^2 + 2x + 3x^3 + 3x^2)/(x(x^2 + x)(x - 1))

Combining like terms in the numerator:

(4x^3 + 2x^2 + x)/(x(x^2 + x)(x - 1))

Therefore, the simplified expression is (4x^3 + 2x^2 + x)/(x(x^2 + x)(x - 1)).

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Answer 3

To simplify the expression ( \frac{1}{x} - \frac{2}{x^2 + x} + \frac{3}{x^3 - x^2} ), we need to find a common denominator for all three fractions. The common denominator in this case is ( x(x^2 + x)(x - 1) ).

After finding the common denominator, we can rewrite each fraction with this denominator and then combine them:

[ \frac{1}{x} - \frac{2}{x^2 + x} + \frac{3}{x^3 - x^2} = \frac{x(x - 1)}{x(x^2 + x)(x - 1)} - \frac{2x(x - 1)}{x(x^2 + x)(x - 1)} + \frac{3x^2}{x(x^2 + x)(x - 1)} ]

Now, combine the fractions:

[ \frac{x(x - 1) - 2x(x - 1) + 3x^2}{x(x^2 + x)(x - 1)} = \frac{x^2 - x - 2x^2 + 2x + 3x^2}{x(x^2 + x)(x - 1)} ]

[ = \frac{0x^2 + x}{x(x^2 + x)(x - 1)} = \frac{x}{x(x^2 + x)(x - 1)} ]

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

So, the simplified form of the expression is ( \frac{1}{(x^2 + x)(x - 1)} ).

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Answer from HIX Tutor

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.

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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