$$6~~~~6~~~~6~~~~6 = 1 \Rightarrow \frac{6}{6} \cdot \frac{6}{6} = 1$$ $$6~~~~6~~~~6~~~~6 = 2 \Rightarrow \frac{6}{6} + \frac{6}{6} = 2$$ $$6~~~~6~~~~6~~~~6 = 3 \Rightarrow\frac{6+6+6}{6} = 3$$ $$ \cdots$$ $$6~~~~6~~~~6~~~~6 = 10 \Rightarrow \frac{6!/6}{6 + 6} = 10$$ I can make four $6$ s equal from $1$ to $10$ except $9$.
Please use any math operators but not by putting two number together like $77$. Could it be done? Thanks.
On
The greatest common divisor (gcd) of sum from $6$ to $12$ and $6$ factorial is equal to $9$.
$GCD(\sum_{6}^{6+6},6!)=9$
On
Assuming we're allowed to use the decimal point (which is often allowed in these kinds of puzzles), we have the following: $$\frac{6}{.6}-\frac{6}{6} = 10-1=9.$$ Note that there's nothing special about the number $6$ here: using the decimal point and the usual operations, you can make $9$ starting with four copies of any number you like.
Let me preface this answer by saying it depends on an abuse of notation, so may not be optimal. That said, it's still a neat little trick!
Following the famous idea of Dirac, I'll give a solution that uses $\log$ and $\sqrt[6]{\cdot}$, but can be used to get any number.
Consider the expression: $- \log_6 \log_6 \sqrt[6]6 = -\log_6 1/6 = 1$
Now to achieve the desired result, note that if we keep applying the root function (and this is where we will have to abuse notation to avoid continuously writing sixes), we can take higher roots, and every time we apply a root inside, the output increases by $1$. So the solution to the OP is the special case where we apply the root function 9 times.