Simplify ((x-4)/(x^2-x))/((x-1)/(2x^2+3x+1)-1/(x^2-1))

Solution:

Step:1

Take the reciprocal of the denominator and multiply it with the numerator: $\frac{x-4}{x^2-x}\cdot \frac{1}{\frac{x-1}{2x^2+3x+1}-\frac{1}{x^2-1}}$

Step:2

Extract $x$ from $x^2-x$.

Step:2.1

Extract $x$ from $x^2$: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{x-1}{2x^2+3x+1}-\frac{1}{x^2-1}}$

Step:3

Use the grouping method to factor.

Step:3.1

For $a{x}^2+bx+c$, split the middle term into two terms whose product is $ac=2\cdot 1=2$ and sum is $b=3$.

Step:3.1.1

Extract $3$ from $3x$: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{x-1}{2x^2+3(x)+1}-\frac{1}{x^2-1}}$

Step:3.1.2

Rewrite $3$ as $1+2$: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{x-1}{2x^2+(1+2)x+1}-\frac{1}{x^2-1}}$

Step:3.1.3

Apply the distributive property: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{x-1}{2x^2+x+2x+1}-\frac{1}{x^2-1}}$

Step:3.2

Factor out the greatest common factor from each group.

Step:3.2.1

Group the first two and the last two terms: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{x-1}{(2x^2+x)+(2x+1)}-\frac{1}{x^2-1}}$

Step:3.2.2

Factor out the GCF from each group: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{x-1}{x(2x+1)+(2x+1)}-\frac{1}{x^2-1}}$

Step:4

Simplify the denominator.

Step:4.1

Rewrite $1$ as $1^2$: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{x-1}{(2x+1)(x+1)}-\frac{1}{x^2-1^2}}$

Step:4.2

Factor using the difference of squares formula: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{x-1}{(2x+1)(x+1)}-\frac{1}{(x+1)(x-1)}}$

Step:5

Simplify the denominator.

Step:5.1

Multiply by $\frac{x-1}{x-1}$ to get a common denominator: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{(x-1{)}^2}{(2x+1)(x+1)(x-1)}-\frac{1}{(x+1)(x-1)}}$

Step:5.2

Multiply by $\frac{2x+1}{2x+1}$ to get a common denominator: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{1}{\frac{(x-1{)}^2}{(2x+1)(x+1)(x-1)}-\frac{2x+1}{(2x+1)(x+1)(x-1)}}$

Step:6

Multiply the numerator by the reciprocal of the denominator: $\frac{x-4}{x\cdot (x-1)}\cdot \frac{(2x+1)(x+1)(x-1)}{(x-1{)}^2-(2x+1)}$

Step:7

Cancel the common factors.

Step:7.1

Cancel $x-4$: $\frac{1}{x\cdot (x-1)}\cdot \frac{(2x+1)(x+1)(x-1)}{x}$

Step:7.2

Cancel $x-1$: $\frac{1}{x}\cdot \frac{(2x+1)(x+1)}{x}$

Step:8

Multiply $\frac{1}{x}$ by $\frac{(2x+1)(x+1)}{x}$: $\frac{(2x+1)(x+1)}{x^2}$

Knowledge Notes:

1. Reciprocal Multiplication: When dividing fractions, multiply by the reciprocal of the divisor.

2. Factoring: Extracting common factors from terms to simplify expressions.

3. Grouping Method: A factoring technique that involves rearranging terms and factoring in groups.

4. Difference of Squares: A formula used to factor expressions in the form of $a^2-b^2=(a+b)(a-b)$.

5. Common Denominator: When combining fractions, a common denominator is required. This often involves multiplying by a form of one to achieve the same denominator.

6. Canceling Common Factors: When the same factor appears in both the numerator and denominator, it can be canceled out.

7. Power Rule: For exponents, $a^m\cdot a^n=a^{m+n}$.

8. Distributive Property: Multiplying a single term across terms within parentheses, such as $a(b+c)=ab+ac$.