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Which expression is a monomial?

A. [tex]$\frac{1}{x}$[/tex]
B. [tex]$3 x^{0.5}$[/tex]
C. [tex]$x + 1$[/tex]
D. [tex]$7$[/tex]


Sagot :

To determine which expressions are monomials, let's first understand what a monomial is. A monomial is a single term algebraic expression that consists of a coefficient and one or more variables raised to non-negative integer powers. Specifically, a monomial:
- Cannot involve division by a variable.
- Cannot have variables with negative or fractional exponents.
- Cannot be a sum or difference of terms (although multiplication is allowed).

Given the expressions:
1. [tex]\(\frac{1}{x}\)[/tex]
2. [tex]\(3x^5\)[/tex]
3. [tex]\(x + 1\)[/tex]
4. [tex]\(7\)[/tex]

Let's analyze each one:

1. [tex]\(\frac{1}{x}\)[/tex]
- This can be rewritten as [tex]\(x^{-1}\)[/tex].
- It has a variable with a negative exponent, which disqualifies it from being a monomial.

2. [tex]\(3x^5\)[/tex]
- This expression has a single term.
- The exponent of [tex]\(x\)[/tex] is a non-negative integer (5).
- There is no division by a variable.
- Hence, [tex]\(3x^5\)[/tex] is a monomial.

3. [tex]\(x + 1\)[/tex]
- This expression has two terms, indicated by the addition operator [tex]\(+\)[/tex].
- Monomials cannot be a sum of terms.
- Hence, [tex]\(x + 1\)[/tex] is not a monomial.

4. [tex]\(7\)[/tex]
- This is a constant term.
- Constants are considered monomials because they can be thought of as [tex]\(7 \times x^0\)[/tex], and [tex]\(x^0 = 1\)[/tex].
- Since the exponent on [tex]\(x\)[/tex] is non-negative (zero) and it fits the definition of a single term, it is a monomial.

Among the given expressions, here are the monomials:

- [tex]\(3x^5\)[/tex]
- [tex]\(7\)[/tex]

Since the question seems to be asking for "which expression" (implying a single choice), we note that [tex]\(3x^5\)[/tex] is more clearly a conventional monomial with both a variable and a coefficient.

So, the correct answer is:
[tex]\[ 3x^5 \][/tex]
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