Write the general term in the expansion of `(x^2-y x)^(12),x!=0`

Explanation

To find the general term in the expansion of \((x^2 - yx)^{12}\), we will use the Binomial Theorem. The Binomial Theorem states that for any positive integer \(n\), the expansion of \((p + q)^n\) can be expressed as:

\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]

where \(T_{r+1}\) is the \((r+1)^{th}\) term, \(p\) and \(q\) are the terms in the binomial, and \(\binom{n}{r}\) is the binomial coefficient.

### Step-by-Step Solution:

1. **Identify \(p\), \(q\), and \(n\)**:
In our case, we have:
- \(p = x^2\)
- \(q = -yx\)
- \(n = 12\)

2. **Write the general term**:
According to the Binomial Theorem, the general term \(T_{r+1}\) in the expansion of \((x^2 - yx)^{12}\) is given by:
\[ T_{r+1} = \binom{12}{r} (x^2)^{12-r} (-yx)^r \]

3. **Simplify the general term**:
Now, we will simplify \(T_{r+1}\):
\[ T_{r+1} = \binom{12}{r} (x^2)^{12-r} (-yx)^r \]
This can be expanded as:
\[ T_{r+1} = \binom{12}{r} (x^{2(12-r)}) (-1)^r (y^r)(x^r) \]
Combining the powers of \(x\):
\[ T_{r+1} = \binom{12}{r} (-1)^r y^r x^{24 - r} \]

4. **Final expression for the general term**:
Thus, the general term in the expansion of \((x^2 - yx)^{12}\) is:
\[ T_{r+1} = \binom{12}{r} (-1)^r y^r x^{24 - r} \]

### Final Answer:
The general term in the expansion of \((x^2 - yx)^{12}\) is:
\[ T_{r+1} = \binom{12}{r} (-1)^r y^r x^{24 - r} \]