To find the value of \( x \) that is less than 51 and makes \( (x, 45, 51) \) a Pythagorean triple, we start with the Pythagorean theorem:
\[ a^2 + b^2 = c^2 \]
In our case, let's assign \( a = x \), \( b = 45 \), and \( c = 51 \). Thus, we have:
\[ x^2 + 45^2 = 51^2 \]
First, we calculate \( 45^2 \) and \( 51^2 \):
\[ 45^2 = 2025 \]
\[ 51^2 = 2601 \]
Now we can substitute these values into the equation:
\[ x^2 + 2025 = 2601 \]
Next, we will isolate \( x^2 \):
\[ x^2 = 2601 - 2025 \]
Calculating the right hand side gives us:
\[ x^2 = 576 \]
Now, we take the square root of both sides to find \( x \):
\[ x = \sqrt{576} = 24 \]
Since 24 is less than 51, it meets the criteria given in the problem.
Thus, the value of \( x \) that makes \( (x, 45, 51) \) a Pythagorean triple is:
\[ \boxed{24} \]