What is the difference? $$ \begin{array}{l}\frac{2 x+5}{x^{2}-3 x}-\frac{3 x+5}{x^{3}-9 x}-\frac{x+1}{x^{2}-9} \ \frac{(x+5)(x+2)}{x^{3}-9 x} \ \frac{(x+5)(x+4)}{x^{3}-9 x} \ \frac{-2 x+11}{x^{3}-12 x-9} \ \frac{3(x+2)}{x^{2}-3 x}\end{array} $$

Question

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We want to compare several rational expressions and see in what way they are different or equivalent. In the list below we have

1) 
$$
\frac{2x+5}{x^{2}-3x}-\frac{3x+5}{x^{3}-9x}-\frac{x+1}{x^{2}-9},
$$

2) 
$$
\frac{(x+5)(x+2)}{x^{3}-9x},
$$

3) 
$$
\frac{(x+5)(x+4)}{x^{3}-9x},
$$

4) 
$$
\frac{-2x+11}{x^{3}-12x-9},
$$

5) 
$$
\frac{3(x+2)}{x^{2}-3x}\,.
$$

Below is a step‐by‐step analysis.

---

**Step 1. Simplify Expression 1**

The denominators in expression (1) can be factorized as
$$
x^{2}-3x = x(x-3),
$$
$$
x^{3}-9x = x(x^{2}-9)=x(x-3)(x+3),
$$
$$
x^{2}-9 = (x-3)(x+3).
$$

Thus, using the common denominator $$\displaystyle x(x-3)(x+3)$$, we rewrite each term:

• The first term becomes
$$
\frac{2x+5}{x(x-3)}=\frac{(2x+5)(x+3)}{x(x-3)(x+3)}\,.
$$

• The second term is already with the common denominator:
$$
\frac{3x+5}{x(x-3)(x+3)}\,.
$$

• The third term becomes
$$
\frac{x+1}{(x-3)(x+3)}=\frac{x(x+1)}{x(x-3)(x+3)}\,.
$$

Now expression (1) becomes
$$
\frac{(2x+5)(x+3) - (3x+5) - x(x+1)}{x(x-3)(x+3)}\,.
$$

Expanding the numerator, we get:

1. Expand $$(2x+5)(x+3)$$:
$$
(2x+5)(x+3)=2x^2+6x+5x+15=2x^2+11x+15\,.
$$

2. Subtract $$(3x+5)$$:
$$
2x^2+11x+15-(3x+5)=2x^2+8x+10\,.
$$

3. Subtract $$x(x+1)=x^2+x$$:
$$
2x^2+8x+10 - (x^2+x)=x^2+7x+10\,.
$$

Notice that the numerator factors nicely:
$$
x^2+7x+10=(x+5)(x+2)\,.
$$

Thus, expression (1) simplifies to
$$
\frac{(x+5)(x+2)}{x(x-3)(x+3)}=\frac{(x+5)(x+2)}{x^3-9x}\,.
$$

---

**Step 2. Compare Expression 1 with Expression 2**

Expression (2) is given as
$$
\frac{(x+5)(x+2)}{x^3-9x}\,.
$$

Since we obtained exactly the same result when simplifying expression (1), we conclude that

$$
\text{Expression (1)}=\text{Expression (2)}\,.
$$

---

**Step 3. Compare with the Remaining Expressions**

1. **Expression (3):**
$$
\frac{(x+5)(x+4)}{x^3-9x}\,.
$$
The numerator $$(x+5)(x+4)$$ is different from $$(x+5)(x+2)$$ (unless for some isolated values of $$x$$ they might coincide). In general, the two expressions are not equivalent.

2. **Expression (4):**
$$
\frac{-2x+11}{x^3-12x-9}\,.
$$
The denominator here is a different cubic:
$$
x^3-12x-9\,,
$$
which, upon checking, factors differently (e.g. $$x=-3$$ is a root, so $$x^3-12x-9=(x+3)(x^2-3x-3)$$). Thus its factorization is not the same as $$x^3-9x=x(x-3)(x+3)$$. Therefore expression (4) is different.

3. **Expression (5):**
$$
\frac{3(x+2)}{x^2-3x}=\frac{3(x+2)}{x(x-3)}\,.
$$
To compare with expression (2) (or (1)), we would need to involve the factor $$(x+3)$$ in the denominator. In fact, if we multiply numerator and denominator of expression (5) by $$(x+3)$$, we obtain
$$
\frac{3(x+2)(x+3)}{x(x-3)(x+3)}=\frac{3(x+2)(x+3)}{x^3-9x}\,.
$$
This is different from
$$
\frac{(x+5)(x+2)}{x^3-9x}\,,
$$
unless $$3(x+3)=(x+5)$$ identically; but that is not true in general. Hence expression (5) is also different.

---

**Conclusion**

- Expression (1) simplifies to $$\frac{(x+5)(x+2)}{x^3-9x}$$ and is equivalent to Expression (2).
- Expression (3) has a different numerator $$(x+5)(x+4)$$ instead of $$(x+5)(x+2)$$.
- Expression (4) has a different denominator.
- Expression (5) is not equivalent to expressions (1) and (2) because its numerator and missing factor in the denominator make it different.

Thus, the only two expressions that are equivalent are (1) and (2); the others differ.
Expressions (1) and (2) are equivalent, while expressions (3), (4), and (5) are different.

What is the difference? \( \begin{array}{l}\frac{2 x+5}{x^{2}-3 x}-\frac{3 x+5}{x^{3}-9 x}-\frac{x+1}{x^{2}-9} \\ \frac{(x+5)(x+2)}{x^{3}-9 x} \\ \frac{(x+5)(x+4)}{x^{3}-9 x} \\ \frac{-2 x+11}{x^{3}-12 x-9} \\ \frac{3(x+2)}{x^{2}-3 x}\end{array} \)

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