To find a rational expression with the domain restriction of \( x \neq -1 \), we need to identify which expression has a denominator that becomes zero when \( x = -1 \).
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Expression 1: \[ \frac{6x^3 + 60}{x} \] The denominator is \( x \), which is zero when \( x = 0 \). So, this does not satisfy the restriction \( x \neq -1 \).
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Expression 2: \[ \frac{\sqrt{x}}{2x + 2} \] The denominator is \( 2x + 2 \). Setting this equal to zero: \[ 2x + 2 = 0 \quad \Rightarrow \quad 2x = -2 \quad \Rightarrow \quad x = -1 \] This expression does have the restriction \( x \neq -1 \).
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Expression 3: \[ \frac{x + 1}{x^3} \] The denominator \( x^3 \) is zero at \( x = 0 \). Thus, it does not satisfy the restriction \( x \neq -1 \).
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Expression 4: \[ \frac{x^2 + 6x + 5}{6x + 6} \] The denominator is \( 6x + 6 \). Setting this equal to zero: \[ 6x + 6 = 0 \quad \Rightarrow \quad 6x = -6 \quad \Rightarrow \quad x = -1 \] This expression also has the restriction \( x \neq -1 \).
Therefore, both Expression 2 and Expression 4 have the domain restriction \( x \neq -1 \).
However, since the question asks for "a rational expression" rather than multiple expressions, mainly Expression 2 and Expression 4 can be considered valid answers based on the criteria.
To summarize, the rational expressions with a restriction of \( x \neq -1 \) are:
- \(\frac{\sqrt{x}}{2x + 2}\) (Expression 2)
- \(\frac{x^2 + 6x + 5}{6x + 6}\) (Expression 4)
If one answer is required, you would choose either of these two.