Math Problem Statement

A=(x+2/2x-4 - x-2/2x+4 - 8/4-x^2) : 4/x-2 * (x^2-2x+3)

Solution

Let's break down the expression step by step and simplify it:

Given:

A=(x+22x4x22x+484x2)÷(4x2×(x22x+3))A = \left( \frac{x + 2}{2x - 4} - \frac{x - 2}{2x + 4} - \frac{8}{4 - x^2} \right) \div \left( \frac{4}{x - 2} \times (x^2 - 2x + 3) \right)

Step 1: Simplifying the first part of the numerator

The first three terms in the numerator are:

  1. x+22x4\frac{x + 2}{2x - 4}
  2. x22x+4\frac{x - 2}{2x + 4}
  3. 84x2\frac{8}{4 - x^2}

We can simplify these terms:

  • Factor the denominators: 2x4=2(x2),2x+4=2(x+2),4x2=(2x)(2+x)2x - 4 = 2(x - 2), \quad 2x + 4 = 2(x + 2), \quad 4 - x^2 = (2 - x)(2 + x)

So, the terms become: x+22(x2)x22(x+2)8(2x)(2+x)\frac{x + 2}{2(x - 2)} - \frac{x - 2}{2(x + 2)} - \frac{8}{(2 - x)(2 + x)}

Notice that 4x2=(2x)(2+x)4 - x^2 = (2 - x)(2 + x), so the third term becomes: 8(x2)(x+2)=8(x2)(x+2)- \frac{8}{-(x - 2)(x + 2)} = \frac{8}{(x - 2)(x + 2)}

So, the expression becomes: x+22(x2)x22(x+2)+8(x2)(x+2)\frac{x + 2}{2(x - 2)} - \frac{x - 2}{2(x + 2)} + \frac{8}{(x - 2)(x + 2)}

Step 2: Combine the terms in the numerator

To combine these terms, let's find a common denominator. The least common denominator (LCD) of 2(x2)2(x - 2), 2(x+2)2(x + 2), and (x2)(x+2)(x - 2)(x + 2) is 2(x2)(x+2)2(x - 2)(x + 2).

Now, rewrite each term with this common denominator:

  • The first term becomes: x+22(x2)=(x+2)(x+2)2(x2)(x+2)=(x+2)22(x2)(x+2)\frac{x + 2}{2(x - 2)} = \frac{(x + 2)(x + 2)}{2(x - 2)(x + 2)} = \frac{(x + 2)^2}{2(x - 2)(x + 2)}

  • The second term becomes: x22(x+2)=(x2)(x2)2(x2)(x+2)=(x2)22(x2)(x+2)- \frac{x - 2}{2(x + 2)} = - \frac{(x - 2)(x - 2)}{2(x - 2)(x + 2)} = \frac{-(x - 2)^2}{2(x - 2)(x + 2)}

  • The third term remains: 8(x2)(x+2)=162(x2)(x+2)\frac{8}{(x - 2)(x + 2)} = \frac{16}{2(x - 2)(x + 2)}

Now combine all the terms: (x+2)2(x2)2+162(x2)(x+2)\frac{(x + 2)^2 - (x - 2)^2 + 16}{2(x - 2)(x + 2)}

Step 3: Simplify the numerator

Expand the squares in the numerator: (x+2)2=x2+4x+4,(x2)2=x24x+4(x + 2)^2 = x^2 + 4x + 4, \quad (x - 2)^2 = x^2 - 4x + 4 So the numerator becomes: (x2+4x+4)(x24x+4)+16(x^2 + 4x + 4) - (x^2 - 4x + 4) + 16 Simplifying: x2+4x+4x2+4x4+16=8x+16x^2 + 4x + 4 - x^2 + 4x - 4 + 16 = 8x + 16

Thus, the numerator is: 8x+168x + 16

Step 4: Simplify the denominator

The denominator is: 4x2×(x22x+3)\frac{4}{x - 2} \times (x^2 - 2x + 3) Multiply the two terms: 4(x22x+3)x2\frac{4(x^2 - 2x + 3)}{x - 2}

Step 5: Final Expression

Now, combine the numerator and denominator: A=8x+162(x2)(x+2)÷4(x22x+3)x2A = \frac{8x + 16}{2(x - 2)(x + 2)} \div \frac{4(x^2 - 2x + 3)}{x - 2}

To divide by a fraction, multiply by its reciprocal: A=8x+162(x2)(x+2)×x24(x22x+3)A = \frac{8x + 16}{2(x - 2)(x + 2)} \times \frac{x - 2}{4(x^2 - 2x + 3)}

Step 6: Simplifying Further

Now, cancel out common terms:

  • The x2x - 2 terms cancel out, leaving: A=8x+162(x+2)×4(x22x+3)A = \frac{8x + 16}{2(x + 2) \times 4(x^2 - 2x + 3)} Simplifying the constants in the denominator: A=8x+168(x+2)(x22x+3)A = \frac{8x + 16}{8(x + 2)(x^2 - 2x + 3)}

Finally, factor the numerator: A=8(x+2)8(x+2)(x22x+3)A = \frac{8(x + 2)}{8(x + 2)(x^2 - 2x + 3)}

Cancel out the common 8(x+2)8(x + 2) terms: A=1x22x+3A = \frac{1}{x^2 - 2x + 3}

Final Answer:

A=1x22x+3A = \frac{1}{x^2 - 2x + 3}

Let me know if you'd like more details or if you have any questions!


Related questions:

  1. How can we factor expressions like x22x+3x^2 - 2x + 3?
  2. What happens when you divide polynomials by fractions?
  3. Can you explain the process of factoring quadratics in detail?
  4. What does the concept of least common denominators mean in simplifying rational expressions?
  5. How does cancellation in fractions work?

Tip: When simplifying complex expressions, break them into smaller steps and carefully factor terms where possible. This helps avoid mistakes and simplifies the process!

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Math Problem Analysis

Mathematical Concepts

Algebra
Rational Expressions
Factoring
Simplification

Formulas

Factoring difference of squares
Simplifying rational expressions
Multiplying and dividing rational expressions

Theorems

Difference of squares theorem
LCD (Least Common Denominator)
Cancellation of common factors in fractions

Suitable Grade Level

Grades 9-12