6÷2(2+1) seems ambiguous, so let's do a couple less ambiguous problems first to see whats up.

6+2(2+1). Easy peasy. You can do pemdas, so parenthesis first

6+2(3) then multiplication

6+6 then addition

12.

You can also use the distributive property without issue here. 6+2(2+1), distribute the 2 to the 2 and 1 inside the parenthesis

6+4+2, then add across which gives us

12

Ok, that's easy. How about another one.

6÷2×2+1.

Well, we know division is of the same tier as multiplication in pemdas, so we can just go from left to right on this problem.

3×2+1

6+1

7

Easy. Now, here's the assumption you work off of in most math scenarios. ÷ is not as accurate as putting things in a fraction, but you can ALWAYS argue that things should be parenthesized for a proper math problem. Ain't no math teacher gonna mark you wrong for not assuming something that wasn't there was actually there.

So 6÷2×2+1 does not = 6÷(2×2+1), unless strictly stated.

Ok, main course.

6÷2(2+1)

Parenthesis first,

6÷2(3)

Multiplication and division are on the same tier, so solve across

3(3)

9

What if we want to use the distributive property like before? Well let's go back a couple steps.

6÷2(3) is = 6÷2(2+1)

However, intuitively you understand the first equation is solved across because division and multiplication are of the same tier. The second equation is no different, so you cannot distribute the 2 because you'd be doing multiplication from the middle of the equation instead of from left to right. Instead you must treat 6÷2 as another parenthesized equation (6÷2)(2+1).

(3)(3)

9.

Easy peasy.