Indicates the corresponding sign:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_β}(5^2-3+6):7\cdot\frac{1}{4}$

For a given problem, whether it involves addition or subtraction **each of the terms that come into play separately**,

this is done **within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are performed before all others,**

__A. We will start with the terms that are on the left in the given problem:__

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2$First, we simplify the terms that are in the denominators in accordance with the order of operations, this is done by calculating the numerical value of the denominator in strength **(this within that we remember that in defining the root as strong, the root itself is strong for everything)** and then perform the operation of division and subtraction:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2 =\\
\frac{1}{\sqrt{16}}\cdot(125+3-4):2^2 =\\
\frac{1}{\sqrt{16}}\cdot124:2^2$Next, we calculate the numerical values of the part that was passed in strength (**practically, if we were to represent the operation of division as broken, this part would have been in the broken position**) and as such the numerical value of the part in strength that in the broken position in the terms, next we perform the operations of multiplication and division:

$\frac{1}{\sqrt{16}}\cdot124:2^2 =\\
\frac{1}{4}\cdot124:4 =\\
\frac{1\cdot124}{4}:4=\\
\frac{\not{124}}{\not{4}}:4=\\
31:4=\\
\frac{31}{4}=\\
7\frac{3}{4}$In the final stages, we performed the multiplication of the number 124 in break, this we did within that we remember **that multiplication in break means multiplication in the broken position**, next we performed the operation of division of the break (by condensing the break) and in the final stage we performed the operation of division in the number 4, this operation resulted in a complete answer, and therefore we marked it as break (a break from understanding, an assumption that the position is greater than the position) and in the continuation we converted the break from understanding to a mixed break, by extracting the wholes (**the answer to the question: "How many times does the division enter the divisor?"**) and adding the remaining division to the divisor,

**We finished simplifying the terms that are on the left in the given problem, we will summarize the simplification stages:**

We received that:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2 =\\
\frac{1}{\sqrt{16}}\cdot124:2^2 =\\
\frac{1\cdot124}{4}:4=\\
7\frac{3}{4}$

__B. We will continue and simplify the terms that are on the right in the given problem:__

$(5^2-3+6):7\cdot\frac{1}{4}$In this part, to simplify the terms within the framework of the order of operations,

In this term, the operation of division of the beginning on terms that are in the denominators, therefore we will first simplify this term,

Let's note that **multiplication precedes addition and subtraction, which precede division and subtraction,** therefore we will start by calculating the numerical value of the part in strength that in this term, next we perform the operations of division and subtraction:

$(5^2-3+6):7\cdot\frac{1}{4} =\\
(25-3+6):7\cdot\frac{1}{4} =\\
28:7\cdot\frac{1}{4}$We will continue and simplify the received term, **noting that between multiplication and division there is no precedence defined in the order of operations, we perform the operations in this term one after the other according to the order from left to right, which is the natural order of operations:**

$28:7\cdot\frac{1}{4} =\\
4\cdot\frac{1}{4} =\\
\frac{4\cdot1}{4}=\\
\frac{\not{4}}{\not{4}}=\\
1$ In the second stage, we performed the multiplication in break, this within that we remember (again) **that multiplication in break means multiplication in the broken position**, in the next stage we performed the operation of division of the break (by condensing the break).

**We finished simplifying the terms that are on the right in the given problem, we will summarize the simplification stages:**

We received that:

$(5^2-3+6):7\cdot\frac{1}{4} =\\
28:7\cdot\frac{1}{4} =\\
\frac{\not{4}}{\not{4}}=\\
1$__We return to the original problem__, and we will present the results of simplifying the terms **that were reported in A and B:**

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_β}(5^2-3+6):7\cdot\frac{1}{4} \\
\downarrow\\
7\frac{3}{4} \text{ }\textcolor{red}{_β}1$__As a result, we receive that:__

$7\frac{3}{4} \text{ }\textcolor{red}{\neq}1$__Therefore, the correct answer here is answer B.__