To determine which of the given rational expressions equals 1 when [tex]\(x = -1\)[/tex], let's evaluate each expression step-by-step at [tex]\(x = -1\)[/tex].
1. For the expression [tex]\(\frac{3x - 2}{x^2 + 5}\)[/tex]:
[tex]\[
\frac{3(-1) - 2}{(-1)^2 + 5} = \frac{-3 - 2}{1 + 5} = \frac{-5}{6}
\][/tex]
The value is [tex]\(\frac{-5}{6}\)[/tex]. This is not equal to 1.
2. For the expression [tex]\(\frac{-4x + 3}{x^2 - 8}\)[/tex]:
[tex]\[
\frac{-4(-1) + 3}{(-1)^2 - 8} = \frac{4 + 3}{1 - 8} = \frac{7}{-7} = -1
\][/tex]
The value is [tex]\(-1\)[/tex]. This is not equal to 1.
3. For the expression [tex]\(\frac{3x - 4}{6x^2}\)[/tex]:
[tex]\[
\frac{3(-1) - 4}{6(-1)^2} = \frac{-3 - 4}{6 \cdot 1} = \frac{-7}{6}
\][/tex]
The value is [tex]\(\frac{-7}{6}\)[/tex]. This is not equal to 1.
4. For the expression [tex]\(\frac{2x^2}{x + 3}\)[/tex]:
[tex]\[
\frac{2(-1)^2}{-1 + 3} = \frac{2 \cdot 1}{2} = \frac{2}{2} = 1
\][/tex]
The value is [tex]\(1\)[/tex]. This is the expression we are looking for.
Therefore, the rational expression that equals 1 when [tex]\(x = -1\)[/tex] is [tex]\(\frac{2x^2}{x + 3}\)[/tex].
