Let the function $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be defined by

$f(x)=\frac{\sin <!--\; \; -->x}{e^{\mathrm{\pi <!--\; \pi \; -->}x}}\frac{(x^{2023}+2024x+2025)}{(x^2-<!--\; -\; -->x+3)}+\frac{2}{e^{\mathrm{\pi <!--\; \pi \; -->}x}}\frac{(x^{2023}+2024x+2025)}{(x^2-<!--\; -\; -->x+3)}$

Then the number of solutions of f(x) = 0 in R is ______.

To find the number of solutions of the equation $f\left(x\right)=0$ in the set of real numbers $ℝ$, let us first write down the expression for the function:
$f\left(x\right)=\frac{\sin x}{e^{\pi x}}·\frac{x^{2023}+2024x+2025}{x^2-x+3}+\frac{2}{e^{\pi x}}·\frac{x^{2023}+2024x+2025}{x^2-x+3}$

We can factor out the common terms from both parts of the sum:
$f\left(x\right)=\frac{x^{2023}+2024x+2025}{e^{\pi x}(x^2-x+3)}(\sin x+2)$

Now, let us analyze each factor to find the roots of the equation $f\left(x\right)=0$:
1. The denominator:
The term $e^{\pi x}$ is strictly positive for all real numbers $x$.
The quadratic term $x^2-x+3$ has a discriminant of $D={(-1)}^2-4\left(1\right)\left(3\right)=1-12=-11<0$. Since the discriminant is negative and the leading coefficient is positive, $x^2-x+3>0$ for all $x\in ℝ$.
Therefore, the denominator is always positive and never zero.

2. The trigonometric factor:
Since the range of the sine function is $[-1,1]$, we have:
$-1\le \sin x\le 1\Rightarrow 1\le \sin x+2\le 3$
Thus, the term $\sin x+2$ is strictly positive and can never be zero.

3. The numerator polynomial:
Since the other components of $f\left(x\right)$ are non-zero, the equation $f\left(x\right)=0$ simplifies to finding the real roots of the polynomial:
$g\left(x\right)=x^{2023}+2024x+2025=0$

Let us find the derivative of $g\left(x\right)$ with respect to $x$:
$g^{\text{'}}\left(x\right)=2023x^{2022}+2024$
Since the exponent $2022$ is even, $x^{2022}\ge 0$ for all real numbers $x$.
Therefore, $g^{\text{'}}\left(x\right)\ge 2024>0$ for all $x\in ℝ$.
Since the derivative is strictly positive, the function $g\left(x\right)$ is strictly increasing on $ℝ$.

Because $g\left(x\right)$ is a continuous polynomial of odd degree (2023), it must have at least one real root. Since it is also strictly increasing, it can cross the x-axis exactly once.
Therefore, the equation $g\left(x\right)=0$ (and consequently $f\left(x\right)=0$) has exactly 1 real solution.