$$\begin{array}{ rlrlrl}
   f_1(x)=&x^2+x & f_2(x)=&2x^2-x & f_3(x)=&x^2 +x \\
   g_1(x)=&x-2 & g_2(x)=&2x \ \  & g_3(x)=&x+2 \\
\end{array}$$
If $h(x)=x$ can be get from $f_i$ and $g_i$ by using only addition, substraction, multiplication defined on those functions where $i\in\{1,2,3\}$, then $F_i=1$. Otherwise, $F_i=0$. What is $(F_1,F_2,F_3)$ ?

$ \textbf{(A)}\ (0,0,0)
\qquad\textbf{(B)}\ (0,0,1)
\qquad\textbf{(C)}\ (0,1,0)
\qquad\textbf{(D)}\ (0,1,1)
\qquad\textbf{(E)}\ \text{None}
$
To determine the values of \( F_i \) for \( i \in \{1, 2, 3\} \), we need to check if \( h(x) = x \) can be obtained using the functions \( f_i(x) \) and \( g_i(x) \) through addition, subtraction, and multiplication.

1. **For \( F_1 \):**
   - Given functions: \( f_1(x) = x^2 + x \) and \( g_1(x) = x - 2 \).
   - We need to check if we can form \( h(x) = x \) using \( f_1(x) \) and \( g_1(x) \).
   - Evaluate at \( x = -1 \):
     \[
     f_1(-1) = (-1)^2 + (-1) = 1 - 1 = 0
     \]
     \[
     g_1(-1) = -1 - 2 = -3
     \]
   - Any combination of \( f_1(x) \) and \( g_1(x) \) will be a multiple of 3 when \( x = -1 \), and thus cannot form \( h(x) = x \).
   - Therefore, \( F_1 = 0 \).

2. **For \( F_2 \):**
   - Given functions: \( f_2(x) = 2x^2 - x \) and \( g_2(x) = 2x \).
   - We need to check if we can form \( h(x) = x \) using \( f_2(x) \) and \( g_2(x) \).
   - Evaluate at \( x = \frac{1}{2} \):
     \[
     f_2\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 - \frac{1}{2} = 2 \cdot \frac{1}{4} - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0
     \]
     \[
     g_2\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} = 1
     \]
   - Any combination of \( f_2(x) \) and \( g_2(x) \) will be an integer when \( x = \frac{1}{2} \), and thus cannot form \( h(x) = x \).
   - Therefore, \( F_2 = 0 \).

3. **For \( F_3 \):**
   - Given functions: \( f_3(x) = x^2 + x \) and \( g_3(x) = x + 2 \).
   - We need to check if we can form \( h(x) = x \) using \( f_3(x) \) and \( g_3(x) \).
   - Consider the expression:
     \[
     g_3(x)^2 - f_3(x) - g_3(x) - g_3(x)
     \]
     \[
     = (x + 2)^2 - (x^2 + x) - (x + 2) - (x + 2)
     \]
     \[
     = x^2 + 4x + 4 - x^2 - x - x - 2 - x - 2
     \]
     \[
     = x^2 + 4x + 4 - x^2 - 3x - 4
     \]
     \[
     = x
     \]
   - Therefore, \( h(x) = x \) can be formed using \( f_3(x) \) and \( g_3(x) \).
   - Thus, \( F_3 = 1 \).

The final answer is \( \boxed{ (0, 0, 1) } \)