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Sagot :
Certainly! Let's simplify the expression step by step.
[tex]\[ \frac{1-1}{1-x+x} - \frac{1}{1+x+x^2} + \frac{2x}{1-x^2+x^4} \][/tex]
First, handle each term separately:
1. [tex]\(\frac{1-1}{1-x+x}\)[/tex]
[tex]\[ \frac{1-1}{1-x+x} = \frac{0}{1-x+x} = 0 \][/tex]
Since the numerator is 0, this entire term simplifies to 0.
2. [tex]\(- \frac{1}{1+x+x^2}\)[/tex]
This term remains unchanged because it cannot be simplified further in its current form.
3. [tex]\(\frac{2x}{1-x^2+x^4}\)[/tex]
Observe that the denominator [tex]\(1 - x^2 + x^4\)[/tex] can be written in a different way:
[tex]\[ 1 - x^2 + x^4 = 1 - x^2 + x^4 = \left(x^2 - x + 1\right)\left(x^2 + x + 1\right) \][/tex]
Therefore,
[tex]\[ \frac{2x}{1-x^2+x^4} = \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
With each term simplified, we now combine them:
[tex]\[ 0 - \frac{1}{1+x+x^2} + \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
This can be written more simply as:
[tex]\[ -\frac{1}{1+x+x^2} + \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
To combine these fractions, find a common denominator. Notice the denominators [tex]\(1 + x + x^2\)[/tex] and [tex]\((x^2 - x + 1)(x^2 + x + 1)\)[/tex].
The common denominator needs to include both terms:
[tex]\[ 1 + x + x^2 \text{ is already part of the factorized form of the other denominator, so the common denominator is \((x^2 - x + 1)(x^2 + x + 1)\)}. \][/tex]
Now,
[tex]\[ -\frac{1}{1 + x + x^2} = -\frac{1 \cdot (x^2 - x + 1)}{(1 + x + x^2)(x^2 - x + 1)} \][/tex]
Which simplifies to:
[tex]\[ -\frac{x^2 - x + 1}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
Therefore our combined fraction becomes:
[tex]\[ -\frac{x^2 - x + 1}{(x^2 - x + 1)(x^2 + x + 1)} + \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
Now, combine the numerators over a common denominator:
[tex]\[ \frac{- (x^2 - x + 1) + 2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
Simplify the numerator:
[tex]\[ = -x^2 + x - 1 + 2x \][/tex]
Combine like terms:
[tex]\[ = -x^2 + 3x - 1 \][/tex]
Putting it all together:
[tex]\[ \frac{-x^2 + 3x - 1}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
So, our simplified expression is:
[tex]\[ \boxed{\frac{-x^2 + 3x - 1}{(x^2 - x + 1)(x^2 + x + 1)}} \][/tex]
[tex]\[ \frac{1-1}{1-x+x} - \frac{1}{1+x+x^2} + \frac{2x}{1-x^2+x^4} \][/tex]
First, handle each term separately:
1. [tex]\(\frac{1-1}{1-x+x}\)[/tex]
[tex]\[ \frac{1-1}{1-x+x} = \frac{0}{1-x+x} = 0 \][/tex]
Since the numerator is 0, this entire term simplifies to 0.
2. [tex]\(- \frac{1}{1+x+x^2}\)[/tex]
This term remains unchanged because it cannot be simplified further in its current form.
3. [tex]\(\frac{2x}{1-x^2+x^4}\)[/tex]
Observe that the denominator [tex]\(1 - x^2 + x^4\)[/tex] can be written in a different way:
[tex]\[ 1 - x^2 + x^4 = 1 - x^2 + x^4 = \left(x^2 - x + 1\right)\left(x^2 + x + 1\right) \][/tex]
Therefore,
[tex]\[ \frac{2x}{1-x^2+x^4} = \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
With each term simplified, we now combine them:
[tex]\[ 0 - \frac{1}{1+x+x^2} + \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
This can be written more simply as:
[tex]\[ -\frac{1}{1+x+x^2} + \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
To combine these fractions, find a common denominator. Notice the denominators [tex]\(1 + x + x^2\)[/tex] and [tex]\((x^2 - x + 1)(x^2 + x + 1)\)[/tex].
The common denominator needs to include both terms:
[tex]\[ 1 + x + x^2 \text{ is already part of the factorized form of the other denominator, so the common denominator is \((x^2 - x + 1)(x^2 + x + 1)\)}. \][/tex]
Now,
[tex]\[ -\frac{1}{1 + x + x^2} = -\frac{1 \cdot (x^2 - x + 1)}{(1 + x + x^2)(x^2 - x + 1)} \][/tex]
Which simplifies to:
[tex]\[ -\frac{x^2 - x + 1}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
Therefore our combined fraction becomes:
[tex]\[ -\frac{x^2 - x + 1}{(x^2 - x + 1)(x^2 + x + 1)} + \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
Now, combine the numerators over a common denominator:
[tex]\[ \frac{- (x^2 - x + 1) + 2x}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
Simplify the numerator:
[tex]\[ = -x^2 + x - 1 + 2x \][/tex]
Combine like terms:
[tex]\[ = -x^2 + 3x - 1 \][/tex]
Putting it all together:
[tex]\[ \frac{-x^2 + 3x - 1}{(x^2 - x + 1)(x^2 + x + 1)} \][/tex]
So, our simplified expression is:
[tex]\[ \boxed{\frac{-x^2 + 3x - 1}{(x^2 - x + 1)(x^2 + x + 1)}} \][/tex]
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