Answer :
To determine which expression is equivalent to [tex]\(\frac{x+3}{x^2-2 x-3} \div \frac{x^2+2 x-3}{x+1}\)[/tex], we will follow the steps of simplifying the given complex fraction.
First, recall the division of fractions rule:
[tex]\[ \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} \][/tex]
So we can rewrite the given expression as:
[tex]\[ \frac{x+3}{x^2-2x-3} \div \frac{x^2+2x-3}{x+1} = \frac{x+3}{x^2-2x-3} \times \frac{x+1}{x^2+2x-3} \][/tex]
Next, let's factorize the quadratic expressions in the denominators and numerators where possible.
1. Factorize [tex]\(x^2-2x-3\)[/tex]:
[tex]\[ x^2-2x-3 = (x-3)(x+1) \][/tex]
2. Factorize [tex]\(x^2+2x-3\)[/tex]:
[tex]\[ x^2+2x-3 = (x+3)(x-1) \][/tex]
Substituting these factorizations into the expression, we get:
[tex]\[ \frac{x+3}{(x-3)(x+1)} \times \frac{x+1}{(x+3)(x-1)} \][/tex]
We can now cancel the common factors in the numerator and the denominator:
- The factor [tex]\(x+3\)[/tex] in the numerator of the first fraction and in the denominator of the second fraction cancels out.
- The factor [tex]\(x+1\)[/tex] in the numerator of the second fraction and in the denominator of the first fraction cancels out.
This simplifies to:
[tex]\[ \frac{1}{(x-3)(x-1)} \][/tex]
Combining the remaining factors in the denominator, we get:
[tex]\[ \frac{1}{x^2 - 4x + 3} \][/tex]
Therefore, the expression simplifies to [tex]\(\frac{1}{x^2 - 4x + 3}\)[/tex]. Looking at the given options, we find that:
B. [tex]\(\frac{1}{x^2-4x+3}\)[/tex]
is the correct answer.
First, recall the division of fractions rule:
[tex]\[ \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} \][/tex]
So we can rewrite the given expression as:
[tex]\[ \frac{x+3}{x^2-2x-3} \div \frac{x^2+2x-3}{x+1} = \frac{x+3}{x^2-2x-3} \times \frac{x+1}{x^2+2x-3} \][/tex]
Next, let's factorize the quadratic expressions in the denominators and numerators where possible.
1. Factorize [tex]\(x^2-2x-3\)[/tex]:
[tex]\[ x^2-2x-3 = (x-3)(x+1) \][/tex]
2. Factorize [tex]\(x^2+2x-3\)[/tex]:
[tex]\[ x^2+2x-3 = (x+3)(x-1) \][/tex]
Substituting these factorizations into the expression, we get:
[tex]\[ \frac{x+3}{(x-3)(x+1)} \times \frac{x+1}{(x+3)(x-1)} \][/tex]
We can now cancel the common factors in the numerator and the denominator:
- The factor [tex]\(x+3\)[/tex] in the numerator of the first fraction and in the denominator of the second fraction cancels out.
- The factor [tex]\(x+1\)[/tex] in the numerator of the second fraction and in the denominator of the first fraction cancels out.
This simplifies to:
[tex]\[ \frac{1}{(x-3)(x-1)} \][/tex]
Combining the remaining factors in the denominator, we get:
[tex]\[ \frac{1}{x^2 - 4x + 3} \][/tex]
Therefore, the expression simplifies to [tex]\(\frac{1}{x^2 - 4x + 3}\)[/tex]. Looking at the given options, we find that:
B. [tex]\(\frac{1}{x^2-4x+3}\)[/tex]
is the correct answer.
