Question 2 (1 point) Perform the indicated operations and simplify. $$ \frac{2-\frac{1}{x}-\frac{2}{x+1}}{\frac{1}{x^{2}+2 x+1}-1} $$ $$ \frac{(2 x+1)(x-1)(x+1)}{-x^{2}(x+2)}, x
eq 0,-1,-2 $$ $$ \frac{(2 x+1)(x-1)(x+1)^{2}}{-x^{2}(x+2)}, x
eq 0,-1,-2 $$ $$ \frac{(2 x+1)(x-1)(x+1)}{-x(x+2)}, x
eq 0,-1,-2 $$ $$ \frac{(2 x+1)(x-1)^{2}(x+1)}{-x^{2}(x+2)}, x
eq 0,-1,-2 $$ $$ \frac{(2 x+1)(x-1)(x+1)}{x^{2}(x+2)}, x
eq 0,-1,-2 $$

Question

Question 2 (1 point) Perform the indicated operations and simplify. $$ \frac{2-\frac{1}{x}-\frac{2}{x+1}}{\frac{1}{x^{2}+2 x+1}-1} $$ $$ \frac{(2 x+1)(x-1)(x+1)}{-x^{2}(x+2)}, x 
eq 0,-1,-2 $$ $$ \frac{(2 x+1)(x-1)(x+1)^{2}}{-x^{2}(x+2)}, x 
eq 0,-1,-2 $$ $$ \frac{(2 x+1)(x-1)(x+1)}{-x(x+2)}, x 
eq 0,-1,-2 $$ $$ \frac{(2 x+1)(x-1)^{2}(x+1)}{-x^{2}(x+2)}, x 
eq 0,-1,-2 $$ $$ \frac{(2 x+1)(x-1)(x+1)}{x^{2}(x+2)}, x 
eq 0,-1,-2 $$

UpStudy AI · Accepted Answer

To simplify the expression

$$
\frac{2 - \frac{1}{x} - \frac{2}{x+1}}{\frac{1}{x^2 + 2x + 1} - 1},
$$

we will first simplify the numerator and the denominator separately.

### Step 1: Simplify the numerator

The numerator is

$$
2 - \frac{1}{x} - \frac{2}{x+1}.
$$

To combine these terms, we need a common denominator. The common denominator for $$x$$ and $$x+1$$ is $$x(x+1)$$. Rewriting each term:

$$
2 = \frac{2x(x+1)}{x(x+1)},
$$
$$
-\frac{1}{x} = -\frac{(x+1)}{x(x+1)} = -\frac{x+1}{x(x+1)},
$$
$$
-\frac{2}{x+1} = -\frac{2x}{x(x+1)}.
$$

Now, combine these:

$$
\frac{2x(x+1) - (x+1) - 2x}{x(x+1)} = \frac{2x^2 + 2x - x - 1 - 2x}{x(x+1)} = \frac{2x^2 - x - 1}{x(x+1)}.
$$

### Step 2: Simplify the denominator

The denominator is

$$
\frac{1}{x^2 + 2x + 1} - 1.
$$

Notice that $$x^2 + 2x + 1 = (x+1)^2$$, so we have:

$$
\frac{1}{(x+1)^2} - 1 = \frac{1 - (x+1)^2}{(x+1)^2}.
$$

Now, simplify $$1 - (x+1)^2$$:

$$
1 - (x^2 + 2x + 1) = 1 - x^2 - 2x - 1 = -x^2 - 2x = -x(x+2).
$$

Thus, the denominator becomes:

$$
\frac{-x(x+2)}{(x+1)^2}.
$$

### Step 3: Combine the numerator and denominator

Now we can combine the simplified numerator and denominator:

$$
\frac{\frac{2x^2 - x - 1}{x(x+1)}}{\frac{-x(x+2)}{(x+1)^2}} = \frac{2x^2 - x - 1}{x(x+1)} \cdot \frac{(x+1)^2}{-x(x+2)}.
$$

This simplifies to:

$$
\frac{(2x^2 - x - 1)(x+1)}{-x^2(x+2)}.
$$

### Step 4: Factor the numerator

Now we need to factor $$2x^2 - x - 1$$. We can factor it as follows:

$$
2x^2 - x - 1 = (2x + 1)(x - 1).
$$

Thus, the numerator becomes:

$$
(2x + 1)(x - 1)(x + 1).
$$

### Final expression

Putting it all together, we have:

$$
\frac{(2x + 1)(x - 1)(x + 1)}{-x^2(x + 2)}.
$$

### Conclusion

The final simplified expression is:

$$
\frac{(2x + 1)(x - 1)(x + 1)}{-x^2(x + 2)}, \quad x \neq 0, -1, -2.
$$

Thus, the correct answer is:

$$
\frac{(2x + 1)(x - 1)(x + 1)}{-x^2(x + 2)}, \quad x \neq 0, -1, -2.
$$
The simplified expression is $$ \frac{(2x + 1)(x - 1)(x + 1)}{-x^2(x + 2)}, x \neq 0, -1, -2 $$.

Question 2 (1 point)
Perform the indicated operations and simplify.

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