https://r2.erweima.ai/mathbot/image/17263733148140snbdqp5.jpg

Let's break down and solve the given expression step by step.

The full expression is:

\[
\frac{2x^2 + 3x}{3x^2 - 6x + xy - 2y} * \frac{x^2 - 4}{18x^3 - 6x^2 y + 2xy^2} \div \frac{2x^2 + 7x + 6}{27x^3 + y^3}
\]

### Step 1: Factorize each term

#### First fraction: 
\[
\frac{2x^2 + 3x}{3x^2 - 6x + xy - 2y}
\]
- Numerator: \(2x^2 + 3x = x(2x + 3)\)
- Denominator: \(3x^2 - 6x + xy - 2y = 3x(x - 2) + y(x - 2) = (x - 2)(3x + y)\)

So, the first fraction simplifies to:
\[
\frac{x(2x + 3)}{(x - 2)(3x + y)}
\]

#### Second fraction:
\[
\frac{x^2 - 4}{18x^3 - 6x^2 y + 2xy^2}
\]
- Numerator: \(x^2 - 4 = (x - 2)(x + 2)\)
- Denominator: \(18x^3 - 6x^2 y + 2xy^2\), factor out \(2x\), which gives:
  \[
  2x(9x^2 - 3xy + y^2)
  \]
  The quadratic term \(9x^2 - 3xy + y^2\) can be written as \((3x - y)^2\).

Thus, the second fraction becomes:
\[
\frac{(x - 2)(x + 2)}{2x(3x - y)^2}
\]

#### Third fraction:
\[
\frac{2x^2 + 7x + 6}{27x^3 + y^3}
\]
- Numerator: \(2x^2 + 7x + 6\) factors as \((2x + 3)(x + 2)\)
- Denominator: \(27x^3 + y^3\) can be factorized as a sum of cubes:
  \[
  27x^3 + y^3 = (3x + y)((3x)^2 - (3x)(y) + y^2) = (3x + y)(9x^2 - 3xy + y^2)
  \]

Thus, the third fraction becomes:
\[
\frac{(2x + 3)(x + 2)}{(3x + y)(9x^2 - 3xy + y^2)}
\]

### Step 2: Simplify the expression

Now, let's multiply the first and second fractions and then divide by the third fraction.

\[
\left(\frac{x(2x + 3)}{(x - 2)(3x + y)} \times \frac{(x - 2)(x + 2)}{2x(3x - y)^2}\right) \div \frac{(2x + 3)(x + 2)}{(3x + y)(9x^2 - 3xy + y^2)}
\]

Cancel out common factors:
- \(x(2x + 3)\) cancels with \((2x + 3)\)
- \((x - 2)\) cancels with \((x - 2)\)
- \((3x + y)\) cancels with \((3x + y)\)
- \((x + 2)\) cancels with \((x + 2)\)

Thus, the simplified expression becomes:
\[
\frac{1}{2x(3x - y)^2} \div \frac{1}{9x^2 - 3xy + y^2}
\]

### Step 3: Simplify the division

Dividing by a fraction is the same as multiplying by its reciprocal, so we multiply by the reciprocal of the third term:
\[
\frac{1}{2x(3x - y)^2} \times \frac{(9x^2 - 3xy + y^2)}{1}
\]

Thus, the final simplified expression is:
\[
\frac{9x^2 - 3xy + y^2}{2x(3x - y)^2}
\]

### Final Answer:
The simplified form of the given expression is:

\[
\frac{9x^2 - 3xy + y^2}{2x(3x - y)^2}
\]

Let me know if you need further clarifications or have any questions! Here are 5 related questions:

1. How do you factor quadratic expressions?
2. What are the steps to simplify complex rational expressions?
3. How do you perform division with algebraic fractions?
4. What is the significance of canceling out common factors in algebra?
5. How do you factor the sum or difference of cubes?

**Tip:** Always factor both the numerator and denominator fully before simplifying any algebraic expression.

Simplify the expression: (2x^2 + 3x) / (3x^2 - 6x + xy - 2y) * (x^2 - 4) / (18x^3 - 6x^2y + 2xy^2) ÷ (2x^2 + 7x + 6) / (27x^3 + y^3)

This solution walks through the simplification of a complex rational algebraic expression involving factoring, rational expressions, and polynomial division. Learn how to simplify step-by-step, with detailed explanations suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation