Evaluate
`(x + 2)/((x + 1)(2x+ 3)) - (2x + 3)/((x + 1)(x + 2)) + (3x + 5)/((2x + 3)(x + 2))`

Options

A. `2x`
B. `-1`
C. `0`
D. `x`

Answer

To evaluate the expression

\[ \frac{x + 2}{(x + 1)(2x + 3)} - \frac{2x + 3}{(x + 1)(x + 2)} + \frac{3x + 5}{(2x + 3)(x + 2)}, \]

we will follow these steps:

### Step 1: Identify the Least Common Multiple (LCM)
The denominators of the fractions are:
1. \((x + 1)(2x + 3)\)
2. \((x + 1)(x + 2)\)
3. \((2x + 3)(x + 2)\)

To find the LCM, we take the highest power of each factor present in the denominators:
- From \((x + 1)\), we take \(x + 1\)
- From \((2x + 3)\), we take \(2x + 3\)
- From \((x + 2)\), we take \(x + 2\)

Thus, the LCM is:

\[ (x + 1)(2x + 3)(x + 2) \]

### Step 2: Rewrite Each Fraction with the Common Denominator
Next, we rewrite each fraction to have the common denominator:

1. For the first fraction:
\[ \frac{x + 2}{(x + 1)(2x + 3)} \cdot \frac{x + 2}{x + 2} = \frac{(x + 2)^2}{(x + 1)(2x + 3)(x + 2)} \]

2. For the second fraction:
\[ \frac{2x + 3}{(x + 1)(x + 2)} \cdot \frac{(2x + 3)}{(2x + 3)} = \frac{(2x + 3)^2}{(x + 1)(2x + 3)(x + 2)} \]

3. For the third fraction:
\[ \frac{3x + 5}{(2x + 3)(x + 2)} \cdot \frac{(x + 1)}{(x + 1)} = \frac{(3x + 5)(x + 1)}{(x + 1)(2x + 3)(x + 2)} \]

### Step 3: Combine the Fractions
Now we can combine the fractions:

\[ \frac{(x + 2)^2 - (2x + 3)^2 + (3x + 5)(x + 1)}{(x + 1)(2x + 3)(x + 2)} \]

### Step 4: Simplify the Numerator
Let's simplify the numerator:

1. Expand \((x + 2)^2\):
\[ (x + 2)^2 = x^2 + 4x + 4 \]

2. Expand \((2x + 3)^2\):
\[ (2x + 3)^2 = 4x^2 + 12x + 9 \]

3. Expand \((3x + 5)(x + 1)\):
\[ (3x + 5)(x + 1) = 3x^2 + 3x + 5x + 5 = 3x^2 + 8x + 5 \]

Now substitute these expansions back into the numerator:

\[ x^2 + 4x + 4 - (4x^2 + 12x + 9) + (3x^2 + 8x + 5) \]

Combine like terms:

\[ (x^2 - 4x^2 + 3x^2) + (4x - 12x + 8x) + (4 - 9 + 5) = 0x^2 + 0x + 0 = 0 \]

### Step 5: Final Result
The numerator simplifies to \(0\), so the entire expression becomes:

\[ \frac{0}{(x + 1)(2x + 3)(x + 2)} = 0 \]

Thus, the value of the expression is:

\[ \boxed{0} \]