jmc

The numerator is equal to $x^{1+2+3+\cdots +9}$. The exponent is the sum of the first 9 consecutive positive integers, so its sum is $\frac{9\cdot 10}{2}=45$. So the numerator is $x^{45}$.

The denominator is equal to $x^{2+4+6+\cdots +12}=x^{2(1+2+3+\cdots +6)}$. The exponent is twice the sum of the first 6 consecutive positive integers, so its sum is $2\cdot \frac{6\cdot 7}{2}=42$. So the denominator is $x^{42}$.

The entire fraction becomes $\frac{x^{45}}{x^{42}}=x^{45-42}=x^3$. Plugging in $x=5$ yields $5^3=\boxed{125}$.