Quadratic trinomial $f(x)$ is allowed to be replaced by one of the trinomials $x^2f(1+\frac{1}{x})$ or $(x-1)^2f(\frac{1}{x-1})$. With the use of these operations, is it possible to go from $x^2+4x+3$ to $x^2+10x+9$?
1. Let's start by analyzing the given quadratic trinomials and the operations allowed. We have:
   \[
   f(x) = x^2 + 4x + 3
   \]
   and we want to determine if it is possible to transform it into:
   \[
   g(x) = x^2 + 10x + 9
   \]

2. The operations allowed are:
   \[
   f(x) \rightarrow x^2 f\left(1 + \frac{1}{x}\right)
   \]
   and
   \[
   f(x) \rightarrow (x-1)^2 f\left(\frac{1}{x-1}\right)
   \]

3. First, let's consider the discriminant of a quadratic trinomial \( ax^2 + bx + c \). The discriminant \( D \) is given by:
   \[
   D = b^2 - 4ac
   \]

4. Calculate the discriminant for \( f(x) = x^2 + 4x + 3 \):
   \[
   D_f = 4^2 - 4 \cdot 1 \cdot 3 = 16 - 12 = 4
   \]

5. Calculate the discriminant for \( g(x) = x^2 + 10x + 9 \):
   \[
   D_g = 10^2 - 4 \cdot 1 \cdot 9 = 100 - 36 = 64
   \]

6. Notice that the discriminants are different:
   \[
   D_f = 4 \quad \text{and} \quad D_g = 64
   \]

7. Now, let's analyze the effect of the allowed operations on the discriminant. Consider the first operation:
   \[
   f(x) \rightarrow x^2 f\left(1 + \frac{1}{x}\right)
   \]
   Let \( f(x) = ax^2 + bx + c \). Then:
   \[
   f\left(1 + \frac{1}{x}\right) = a\left(1 + \frac{1}{x}\right)^2 + b\left(1 + \frac{1}{x}\right) + c
   \]
   Simplifying:
   \[
   f\left(1 + \frac{1}{x}\right) = a\left(1 + \frac{2}{x} + \frac{1}{x^2}\right) + b\left(1 + \frac{1}{x}\right) + c
   \]
   \[
   = a + \frac{2a}{x} + \frac{a}{x^2} + b + \frac{b}{x} + c
   \]
   \[
   = a + b + c + \left(\frac{2a + b}{x}\right) + \left(\frac{a}{x^2}\right)
   \]
   Multiplying by \( x^2 \):
   \[
   x^2 f\left(1 + \frac{1}{x}\right) = a x^2 + (2a + b)x + (a + b + c)
   \]

8. The new quadratic trinomial is:
   \[
   a x^2 + (2a + b)x + (a + b + c)
   \]
   The discriminant of this new trinomial is:
   \[
   D' = (2a + b)^2 - 4a(a + b + c)
   \]
   Simplifying:
   \[
   D' = 4a^2 + 4ab + b^2 - 4a^2 - 4ab - 4ac = b^2 - 4ac
   \]
   Thus, the discriminant remains invariant under this operation.

9. Consider the second operation:
   \[
   f(x) \rightarrow (x-1)^2 f\left(\frac{1}{x-1}\right)
   \]
   Let \( f(x) = ax^2 + bx + c \). Then:
   \[
   f\left(\frac{1}{x-1}\right) = a\left(\frac{1}{x-1}\right)^2 + b\left(\frac{1}{x-1}\right) + c
   \]
   Simplifying:
   \[
   f\left(\frac{1}{x-1}\right) = \frac{a}{(x-1)^2} + \frac{b}{x-1} + c
   \]
   Multiplying by \( (x-1)^2 \):
   \[
   (x-1)^2 f\left(\frac{1}{x-1}\right) = a + b(x-1) + c(x-1)^2
   \]
   \[
   = a + bx - b + cx^2 - 2cx + c
   \]
   \[
   = cx^2 + (b - 2c)x + (a - b + c)
   \]

10. The new quadratic trinomial is:
    \[
    cx^2 + (b - 2c)x + (a - b + c)
    \]
    The discriminant of this new trinomial is:
    \[
    D'' = (b - 2c)^2 - 4c(a - b + c)
    \]
    Simplifying:
    \[
    D'' = b^2 - 4bc + 4c^2 - 4ac + 4bc - 4c^2 = b^2 - 4ac
    \]
    Thus, the discriminant remains invariant under this operation as well.

11. Since the discriminant is invariant under both operations, and the discriminants of \( f(x) \) and \( g(x) \) are different, it is impossible to transform \( f(x) \) into \( g(x) \) using the allowed operations.

The final answer is \( \boxed{ \text{isn't possible to go from}\ x^2+4x+3\ \text{to}\ x^2+10x+9 } \)