Simplify - 6: A math worksheet with 15 algebraic expressions to simplify, provided by NallPro Education Centre.

Problem: Simplify the given expressions and show that they are equal to the provided results.

We will solve each problem step by step. Here is the breakdown:

---

#### (1) Simplify:
\[
\frac{x + 2xy}{3x^2} \div \frac{2y + 1}{6x} = 2
\]

Solution:
1. Rewrite the division as multiplication by the reciprocal:
\[
\frac{x + 2xy}{3x^2} \div \frac{2y + 1}{6x} = \frac{x + 2xy}{3x^2} \cdot \frac{6x}{2y + 1}
\]

2. Simplify the numerator and denominator:
\[
\frac{x + 2xy}{3x^2} \cdot \frac{6x}{2y + 1} = \frac{x(1 + 2y)}{3x^2} \cdot \frac{6x}{2y + 1}
\]

3. Cancel out common factors:
\[
\frac{x(1 + 2y)}{3x^2} \cdot \frac{6x}{2y + 1} = \frac{1 + 2y}{3x} \cdot \frac{6x}{2y + 1}
\]

4. Notice that \(1 + 2y = 2y + 1\), so:
\[
\frac{1 + 2y}{3x} \cdot \frac{6x}{2y + 1} = \frac{6x}{3x} = 2
\]

Thus, the expression simplifies to:
\[
\boxed{2}
\]

---

#### (2) Simplify:
\[
\frac{9 - x^2}{x^4 + 6x^3} \div \frac{x^3 - 2x^2 - 3x}{x^2 + 7x + 6} = -\frac{3 + x}{x^4}
\]

Solution:
1. Rewrite the division as multiplication by the reciprocal:
\[
\frac{9 - x^2}{x^4 + 6x^3} \div \frac{x^3 - 2x^2 - 3x}{x^2 + 7x + 6} = \frac{9 - x^2}{x^4 + 6x^3} \cdot \frac{x^2 + 7x + 6}{x^3 - 2x^2 - 3x}
\]

2. Factor each term:
- \(9 - x^2 = (3 - x)(3 + x)\)
- \(x^4 + 6x^3 = x^3(x + 6)\)
- \(x^3 - 2x^2 - 3x = x(x^2 - 2x - 3) = x(x - 3)(x + 1)\)
- \(x^2 + 7x + 6 = (x + 1)(x + 6)\)

3. Substitute the factored forms:
\[
\frac{(3 - x)(3 + x)}{x^3(x + 6)} \cdot \frac{(x + 1)(x + 6)}{x(x - 3)(x + 1)}
\]

4. Cancel out common factors:
- \(x + 6\) in the numerator and denominator
- \(x + 1\) in the numerator and denominator
- \(3 - x\) in the numerator and \(x - 3\) in the denominator (note that \(3 - x = -(x - 3)\))

\[
\frac{(3 - x)(3 + x)}{x^3(x + 6)} \cdot \frac{(x + 1)(x + 6)}{x(x - 3)(x + 1)} = \frac{-(3 + x)}{x^4}
\]

5. Simplify:
\[
\frac{-(3 + x)}{x^4} = -\frac{3 + x}{x^4}
\]

Thus, the expression simplifies to:
\[
\boxed{-\frac{3 + x}{x^4}}
\]

---

#### (3) Simplify:
\[
\frac{1}{x} + \frac{1}{y} = xy
\]

Solution:
1. Combine the fractions on the left-hand side:
\[
\frac{1}{x} + \frac{1}{y} = \frac{y + x}{xy}
\]

2. The equation becomes:
\[
\frac{x + y}{xy} = xy
\]

3. Multiply both sides by \(xy\) to clear the denominator:
\[
x + y = (xy)^2
\]

This equation is not generally true for all \(x\) and \(y\). It seems there might be a misunderstanding in the problem statement. However, if we assume the goal is to simplify the left-hand side, the simplified form is:
\[
\boxed{\frac{x + y}{xy}}
\]

---

#### (4) Simplify:
\[
\frac{x + y^2}{x^2} + \frac{x - 1}{x} - 1 = \frac{y^2}{x^2}
\]

Solution:
1. Combine the terms on the left-hand side:
\[
\frac{x + y^2}{x^2} + \frac{x - 1}{x} - 1
\]

2. Rewrite \(\frac{x - 1}{x}\) with a common denominator \(x^2\):
\[
\frac{x - 1}{x} = \frac{x(x - 1)}{x^2} = \frac{x^2 - x}{x^2}
\]

3. Rewrite \(-1\) with a common denominator \(x^2\):
\[
-1 = \frac{-x^2}{x^2}
\]

4. Combine all terms:
\[
\frac{x + y^2}{x^2} + \frac{x^2 - x}{x^2} + \frac{-x^2}{x^2} = \frac{x + y^2 + x^2 - x - x^2}{x^2}
\]

5. Simplify the numerator:
\[
x + y^2 + x^2 - x - x^2 = y^2
\]

6. The expression becomes:
\[
\frac{y^2}{x^2}
\]

Thus, the expression simplifies to:
\[
\boxed{\frac{y^2}{x^2}}
\]

---

#### (5) Simplify:
\[
\frac{1}{x + 2} + \frac{1}{x - 2} - \frac{x}{x^2 - 4} = \frac{x}{x^2 - 4}
\]

Solution:
1. Notice that \(x^2 - 4 = (x + 2)(x - 2)\). Rewrite the fractions with a common denominator:
\[
\frac{1}{x + 2} + \frac{1}{x - 2} - \frac{x}{x^2 - 4} = \frac{(x - 2) + (x + 2) - x}{(x + 2)(x - 2)}
\]

2. Simplify the numerator:
\[
(x - 2) + (x + 2) - x = x - 2 + x + 2 - x = x
\]

3. The expression becomes:
\[
\frac{x}{(x + 2)(x - 2)} = \frac{x}{x^2 - 4}
\]

Thus, the expression simplifies to:
\[
\boxed{\frac{x}{x^2 - 4}}
\]

---

#### (6) Simplify:
\[
\frac{2 + \frac{1}{x}}{2x^2 + x} = \frac{1}{x^2}
\]

Solution:
1. Simplify the numerator:
\[
2 + \frac{1}{x} = \frac{2x + 1}{x}
\]

2. Substitute back into the expression:
\[
\frac{2 + \frac{1}{x}}{2x^2 + x} = \frac{\frac{2x + 1}{x}}{2x^2 + x}
\]

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

4. Factor the denominator \(2x^2 + x\):
\[
2x^2 + x = x(2x + 1)
\]

5. Substitute and cancel out \(2x + 1\):
\[
\frac{2x + 1}{x(2x + 1)} = \frac{1}{x}
\]

6. The expression becomes:
\[
\frac{1}{x} \cdot \frac{1}{x} = \frac{1}{x^2}
\]

Thus, the expression simplifies to:
\[
\boxed{\frac{1}{x^2}}
\]

---

Final Answer:

\[
\boxed{2, -\frac{3 + x}{x^4}, \frac{x + y}{xy}, \frac{y^2}{x^2}, \frac{x}{x^2 - 4}, \frac{1}{x^2}}
\]