🍪+🍩+🍪=22 🍫+🍫+🍪=28 🍩+🍩🍩+🍫🍫=38 🍩🍩×🍪-🍫🍫=?

Let's denote the cookies, donuts, and chocolates as follows:

• \( c \) = cookie (🍪)
• \( d \) = donut (🍩)
• \( h \) = chocolate (🍫)

From the equations, we can set up the following system:

1. \( c + d + c = 22 \) → \( 2c + d = 22 \) (Equation 1)
2. \( h + h + c = 28 \) → \( 2h + c = 28 \) (Equation 2)
3. \( d + d + d + h + h = 38 \) → \( 3d + 2h = 38 \) (Equation 3)

Now, let's solve this system step by step.

From Equation 1: \[ d = 22 - 2c \quad \text{(Substituting into other equations)} \]

Substituting \( d \) into Equation 3: \[ 3(22 - 2c) + 2h = 38 \] \[ 66 - 6c + 2h = 38 \] \[ 2h = 38 - 66 + 6c \] \[ 2h = 6c - 28 \quad \Rightarrow \quad h = 3c - 14 \quad \text{(Equation 4)} \]

Substituting Equation 4 into Equation 2: \[ 2(3c - 14) + c = 28 \] \[ 6c - 28 + c = 28 \] \[ 7c - 28 = 28 \] \[ 7c = 56 \quad \Rightarrow \quad c = 8 \]

Now substituting \( c = 8 \) back into Equation 1: \[ 2(8) + d = 22 \] \[ 16 + d = 22 \quad \Rightarrow \quad d = 6 \]

Now substitute \( c = 8 \) into Equation 4 to find \( h \): \[ h = 3(8) - 14 \] \[ h = 24 - 14 \quad \Rightarrow \quad h = 10 \]

Now we have:

• \( c = 8 \)
• \( d = 6 \)
• \( h = 10 \)

Next, we need to calculate \( d d \times c - h h \): \[ d d \times c = 6 \times 6 \times 8 = 36 \times 8 = 288 \] \[ h h = 10 \times 10 = 100 \] Now we can find: \[ d d \times c - h h = 288 - 100 = 188 \]

Thus, the answer is: \[ \boxed{188} \]