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To simplify the given expression \(\frac{1}{x^2-5x+6} - \frac{2}{x^2-4x+3} - \frac{1}{x^2-3x+2}\), let's follow the steps below:
### Step 1: Factorize each denominator
First, we need to factorize each of the quadratic expressions in the denominators.
1. Factorize \(x^2 - 5x + 6\):
[tex]\[ x^2 - 5x + 6 = (x - 2)(x - 3) \][/tex]
2. Factorize \(x^2 - 4x + 3\):
[tex]\[ x^2 - 4x + 3 = (x - 1)(x - 3) \][/tex]
3. Factorize \(x^2 - 3x + 2\):
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
### Step 2: Rewrite the original expression using factored forms
Now, let's rewrite the original expression using the factored forms of the denominators:
[tex]\[ \frac{1}{(x-2)(x-3)} - \frac{2}{(x-1)(x-3)} - \frac{1}{(x-1)(x-2)} \][/tex]
### Step 3: Find a common denominator
The common denominator for these fractions would be the product of the unique factors appearing in each term:
[tex]\[ \text{Common Denominator} = (x-1)(x-2)(x-3) \][/tex]
### Step 4: Rewrite each fraction with the common denominator
Convert each fraction to have the common denominator:
[tex]\[ \frac{1}{(x-2)(x-3)} \cdot \frac{(x-1)}{(x-1)} = \frac{x-1}{(x-1)(x-2)(x-3)} \][/tex]
[tex]\[ \frac{2}{(x-1)(x-3)} \cdot \frac{(x-2)}{(x-2)} = \frac{2(x-2)}{(x-1)(x-2)(x-3)} \][/tex]
[tex]\[ \frac{1}{(x-1)(x-2)} \cdot \frac{(x-3)}{(x-3)} = \frac{x-3}{(x-1)(x-2)(x-3)} \][/tex]
### Step 5: Combine the fractions
Now combine the fractions over the common denominator:
[tex]\[ \frac{(x-1) - 2(x-2) - (x-3)}{(x-1)(x-2)(x-3)} \][/tex]
### Step 6: Simplify the numerator
Simplify the numerator by combining like terms:
[tex]\[ (x-1) - 2(x-2) - (x-3) \\ = x - 1 - 2x + 4 - x + 3 \\ = -2x + x + x + 4 - 1 - 3 \\ = -2 \][/tex]
So the numerator simplifies to \(-2\).
### Step 7: Write the final simplified expression
Putting it all together, the simplified form of the given expression is:
[tex]\[ \frac{-2}{(x-1)(x-2)(x-3)} \][/tex]
Finally, notice that \((x-1)(x-2)(x-3)\) is equivalent to \(x^2-3x+2\):
[tex]\[ \frac{-2}{x^2 - 3x + 2} \][/tex]
Therefore, the simplified form of the expression \(\frac{1}{x^2-5x+6} - \frac{2}{x^2-4x+3} - \frac{1}{x^2-3x+2}\) is:
[tex]\[ \boxed{\frac{-2}{x^2 - 3x + 2}} \][/tex]
### Step 1: Factorize each denominator
First, we need to factorize each of the quadratic expressions in the denominators.
1. Factorize \(x^2 - 5x + 6\):
[tex]\[ x^2 - 5x + 6 = (x - 2)(x - 3) \][/tex]
2. Factorize \(x^2 - 4x + 3\):
[tex]\[ x^2 - 4x + 3 = (x - 1)(x - 3) \][/tex]
3. Factorize \(x^2 - 3x + 2\):
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
### Step 2: Rewrite the original expression using factored forms
Now, let's rewrite the original expression using the factored forms of the denominators:
[tex]\[ \frac{1}{(x-2)(x-3)} - \frac{2}{(x-1)(x-3)} - \frac{1}{(x-1)(x-2)} \][/tex]
### Step 3: Find a common denominator
The common denominator for these fractions would be the product of the unique factors appearing in each term:
[tex]\[ \text{Common Denominator} = (x-1)(x-2)(x-3) \][/tex]
### Step 4: Rewrite each fraction with the common denominator
Convert each fraction to have the common denominator:
[tex]\[ \frac{1}{(x-2)(x-3)} \cdot \frac{(x-1)}{(x-1)} = \frac{x-1}{(x-1)(x-2)(x-3)} \][/tex]
[tex]\[ \frac{2}{(x-1)(x-3)} \cdot \frac{(x-2)}{(x-2)} = \frac{2(x-2)}{(x-1)(x-2)(x-3)} \][/tex]
[tex]\[ \frac{1}{(x-1)(x-2)} \cdot \frac{(x-3)}{(x-3)} = \frac{x-3}{(x-1)(x-2)(x-3)} \][/tex]
### Step 5: Combine the fractions
Now combine the fractions over the common denominator:
[tex]\[ \frac{(x-1) - 2(x-2) - (x-3)}{(x-1)(x-2)(x-3)} \][/tex]
### Step 6: Simplify the numerator
Simplify the numerator by combining like terms:
[tex]\[ (x-1) - 2(x-2) - (x-3) \\ = x - 1 - 2x + 4 - x + 3 \\ = -2x + x + x + 4 - 1 - 3 \\ = -2 \][/tex]
So the numerator simplifies to \(-2\).
### Step 7: Write the final simplified expression
Putting it all together, the simplified form of the given expression is:
[tex]\[ \frac{-2}{(x-1)(x-2)(x-3)} \][/tex]
Finally, notice that \((x-1)(x-2)(x-3)\) is equivalent to \(x^2-3x+2\):
[tex]\[ \frac{-2}{x^2 - 3x + 2} \][/tex]
Therefore, the simplified form of the expression \(\frac{1}{x^2-5x+6} - \frac{2}{x^2-4x+3} - \frac{1}{x^2-3x+2}\) is:
[tex]\[ \boxed{\frac{-2}{x^2 - 3x + 2}} \][/tex]
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