Parentheses in advanced Order of Operations: Identify the greater value

Let's solve exercise A:

$8\times8=64$

Let's solve exercise B:

$10\times6=60$

Let's solve exercise C:

$12\times5=60$

Let's solve exercise D:

$15\times4=60$

To solve a problem given in division or multiplication each of the terms that appear in its expression 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 all others,

A. We will start with the terms that appear on the left in the given problem:

$\frac{1}{9}\cdot((4^2-3\cdot2):2+4)$First, we simplify the terms in the parentheses (the numerators) that multiply the fraction according to the order of operations, noting that the term in the parentheses includes within it an operation of division of the term in the parentheses (the denominators), therefore, we will start simply with this term, in this term a subtraction operation is performed between terms that strengthens the division between terms, therefore the calculation of its numerical value is carried out first followed by the multiplication of the terms and continue to perform the subtraction operation:

$\frac{1}{9}\cdot((4^2-3\cdot2):2+4) =\\ \frac{1}{9}\cdot((16-3\cdot2):2+4) =\\ \frac{1}{9}\cdot((16-6):2+4) =\\ \frac{1}{9}\cdot(10:2+4)\\$Note that there is no prohibition to calculate their numerical values of the term that strengthens as in the term in the parentheses in contrast to the multiplication that in the term in the parentheses, this from a concept that breaks in separate terms, also for the sake of good order we performed this step after step,

We continue simply with the terms in the parentheses that were left, we remember that division precedes subtraction and therefore we will start from calculating the outcome of the multiplication in the term, in the next step the division is performed and finally the multiplication in the break that multiplies the term in the parentheses:

$\frac{1}{9}\cdot(10:2+4)\\ \frac{1}{9}\cdot(5+4)=\\ \frac{1}{9}\cdot9=\\ \frac{1\cdot9}{9}=\\ \frac{\not{9}}{\not{9}}=\\ 1$In the last steps we performed the multiplication of the number 9 in the break, this we did while we remember that the multiplication in the break means the multiplication in the amount of the break,

We finished simply with the terms that appear on the left in the given problem, we will summarize the steps of the simplification:

We received that:

$\frac{1}{9}\cdot((4^2-3\cdot2):2+4) =\\ \frac{1}{9}\cdot(10:2+4)\\ \frac{1}{9}\cdot(5+4)=\\ \frac{9}{9}=\\ 1$

B. We will continue and simplify the terms that appear on the right in the given problem:

$4^2-(3\cdot2:2+4)\cdot\frac{1}{7}$In this part to be done in the first part we simplify the terms within the framework of the order of operations,

In this term a multiplication operation is performed on the term in the parentheses, therefore, we will simplify first this term, we remember that multiplication and division precede subtraction, therefore, we will calculate first the numerical values of the first term from the left in this term, noting that the concept that between multiplication and division there is no predetermined precedence in the order of operations, the operations in this term are performed one after the other according to the order from left to right, which is the natural order of operations, in contrast we will calculate the numerical values of the term that strengthens:

$4^2-(3\cdot2:2+4)\cdot\frac{1}{7} =\\ 16-(6:2+4)\cdot\frac{1}{7} =\\ 16-(3+4)\cdot\frac{1}{7} =\\ 16-7\cdot\frac{1}{7}\$We will continue and perform the multiplication in the break, this within that we remember that the multiplication in the break means the multiplication in the amount of the break, in the next step the division operation of the break (by the compression of the break) is performed and in the last step the remaining subtraction operation, this in accordance with the order of operations:

$16-7\cdot\frac{1}{7}=\\ 16-\frac{7\cdot1}{7}=\\ 16-\frac{\not{7}}{\not{7}}=\\ 16-1=\\ 15$We finished simply with the terms that appear on the right in the given problem, we will summarize the steps of the simplification:

We received that:

$4^2-(3\cdot2:2+4)\cdot\frac{1}{7} =\\ 16-(3+4)\cdot\frac{1}{7} =\\ 16-1=\\ 15$We will return to the original problem, and we will present the outcomes of the simplifications that were reported in A and B:

$\frac{1}{9}\cdot((4^2-3\cdot2):2+4)\textcolor{red}{☐}4^2-(3\cdot2:2+4)\cdot\frac{1}{7} \\ \downarrow\\ 1 \textcolor{red}{☐}15$As a result that is established that:

$1 \text{ }\textcolor{red}{\neq}15$Therefore, the correct answer here is answer B.

According to the given problem, whether it is discussed in addition or in subtraction each of the terms that comes in its turn,

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 all others,

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

$\frac{1}{9}\cdot\big((5^2-8:2):7\big)^2$Begin by simplifying the terms that are in parentheses (the denominators), which are prioritized in strength, the multiplication divides the break in accordance with the order of operations mentioned, note that the terms in the denominators include within them the operation of division part of the term in parentheses (the numerators), therefore we will start simply with this term, in this term the operation of subtraction is performed between a numerator of greater strength to a numerator of lesser strength, therefore the beginning will execute the calculation of their numerical values which in the course of being executed the operation of division and continues to perform the operation of subtraction, in the last stage the operation of division part on the term in parentheses is executed:

$\frac{1}{9}\cdot\big((5^2-8:2):7\big)^2= \\ \frac{1}{9}\cdot\big((25-8:2):7\big)^2 = \\ \frac{1}{9}\cdot\big((25-4):7\big)^2= \\ \frac{1}{9}\cdot\big(21:7\big)^2=\\ \frac{1}{9}\cdot3^2=\\$note that there is no prohibition to calculate their numerical values of the numerator that is stronger which in the term in parentheses in comparison to calculate the numerators which in the term in parentheses, this means that breaking in separate numerators, even for the sake of good order we performed this stage after stage,

We will continue simply the term we received in the last stage, we remember that multiplication precedes division and therefore we will start from calculating their numerical values of the numerator that is stronger, in the next stage the multiplication in the break is executed, and at the end the operation of division (by summarizing the break received):

$\frac{1}{9}\cdot3^2=\\ \frac{1}{9}\cdot9=\\ \frac{1\cdot9}{9}=\\ \frac{\not{9}}{\not{9}}=\\ 1$In the last stages we performed the multiplication of the number 9 in the break, this we performed within that we remember that the multiplication in the break means the multiplication in the amount of the break,

We finished simply the term that is on the left in the given problem, we will summarize the stages of the simplification:

We received that:

$\frac{1}{9}\cdot\big((5^2-8:2):7\big)^2= \\ \frac{1}{9}\cdot\big((25-4):7\big)^2= \\ \frac{1}{9}\cdot3^2=\\ 1$

B. We will continue and simplify the term that is on the right in the given problem:

$\big((-49)\cdot(-2)+3:3\big)\cdot\frac{1}{99}$In this part to perform in the first part simplify the term within the framework on the order of operations,

In this term the operation of division part on the term in parentheses is established, therefore we will simplify starting with this term, we remember that multiply and division precede addition, therefore we will calculate starting their numerical values of the multiplication in the term (this within that we remember that multiplying a series of numbers in a series of numbers gives a mandatory result) and in comparison we will calculate the result of the operation of division in the term, in the next stage the operation of addition in the term is executed:

$\big((-49)\cdot(-2)+3:3\big)\cdot\frac{1}{99} \\ \big(98+1\big)\cdot\frac{1}{99} =\\ 99\cdot\frac{1}{99} \\$We will continue and execute the multiplication in the break, this within that we remember that the multiplication in the break means the multiplication in the amount of the break, in the next stage the operation of division of the break is executed (by summarizing the break):

$99\cdot\frac{1}{99} \\ \frac{ 99\cdot1}{99} \\ \frac{\not{99}}{\not{99}}=\\ 1$We finished simply the term that is on the right in the given problem, we will summarize the stages of the simplification:

We received that:

$\big((-49)\cdot(-2)+3:3\big)\cdot\frac{1}{99} \\ 99\cdot\frac{1}{99} =\\ 1$We will return now to the original problem, and we will present the results of the simplification of the terms that were reported in A and in B:

$\frac{1}{9}\cdot\big((5^2-8:2):7\big)^2\textcolor{red}{☐}\big((-49)\cdot(-2)+3:3\big)\cdot\frac{1}{99} \\ \downarrow\\ 1\textcolor{red}{☐}1$As a result that is established that:

$1 \text{ }\textcolor{red}{=}1$Therefore the correct answer here is answer B.

For solving a problem involving addition or subtraction each of the terms that come into play is treated 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 term that appears on the left in the given problem:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)$In this term, a division operation is established between two terms that in their sum give a specific result, we take the term that in their sum gives the left, remembering that multiplication precedes subtraction, therefore the multiplication in these terms is performed first and then the subtraction operation, in contrast- the term that in their sum gives the right (the denominators) is considered as a multiplication of a number by the term that in their sum (the numerators) therefore the multiplication in this term, this is within that we remember that multiplication precedes division and that the root cause (the definition of the root as strong) is strong for everything, therefore we will calculate its numerical value and then perform the division operation that in this term:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big) =\\ (36-12):\big((4+2)\cdot\frac{1}{6}\big) =\\ 24:\big(6\cdot\frac{1}{6}\big) \\$We will continue and note that in the term that was received in the last stage the multiplication operation is found in the terms and accordingly precedence is given to the division operation that precedes them,the multiplication is performed within that we remember that the multiplication in the break means the multiplication by the break, in continuation the division operation of the break is performed, this by summarizing:

$24:\frac{6\cdot1}{6}=\\ 24:\frac{\not{6}}{\not{6}}=\\ 24:1=\\ 24$In the last stage we performed the remaining division operation, this within that we remember that dividing any number by the number 1 will yield the number itself,

$$We will conclude the simplification of the term that appears on the left in the given problem, we will summarize the simplification stages:

We received that:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big) =\\ 24:\big(6\cdot\frac{1}{6}\big)= \\ 24$

B. We will continue and simplify the term that appears on the right in the given problem:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13}$In this part, in the first section, we simplify the term within the framework of the order of operations,

In this term, a division operation is established between two terms that in their sum, in this part, in the first section, we simplify that two terms in contrast, the term that in their sum gives the left is simplified within performing the division and subtraction operations, in contrast we simplify the term that in their sum gives the right, given that multiplication and division precede subtraction we start from the second simplification stage in these terms , and given that the order of operations does not define precedence to one of the multiplication or division operations is performed one after the other according to the order from left to right (which is the natural order of operations) , in continuation we will calculate the result of the subtraction operation that in these terms:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} =\\ 17:(17-20:5)\cdot\frac{1}{13} =\\ 17:(17-4)\cdot\frac{1}{13} =\\ 17:13\cdot\frac{1}{13} \\$We will continue and simplify the term that was received in the last stage, in this part, first the division and multiplication operations are performed one after the other from left to right:

$17:13\cdot\frac{1}{13} =\\ \frac{17}{13}\cdot\frac{1}{13} =\\ \frac{17\cdot1}{13\cdot13}=\\ \frac{17}{13^2}$In the first stage, given that the result of the division operation is a result that is not whole we mark it as a break (a break from above- given that the numerator is larger than the denominator) in continuation we perform the multiplication of the breaks within that we remember that when we multiply two breaks we multiply numerator by numerator and denominator by denominator and keep the essence of the original break.

We will conclude the simplification of the term that appears on the right in the given problem, we will summarize the simplification stages:

We received that:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} =\\ 17:(17-20:5)\cdot\frac{1}{13} =\\ 17:13\cdot\frac{1}{13} \\ \frac{17}{13}\cdot\frac{1}{13} =\\ \frac{17}{13^2}$We will return to the original problem, and we will present the results of the simplification of the terms that were reported in A and B:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)\textcolor{red}{☐}(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} \downarrow\\ 24 \textcolor{red}{☐}\frac{17}{13^2}$As a result, the correct answer here is answer B.

According to the given problem, whether it's an addition or subtraction each of the digits that come up 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 done before division and subtraction, and that the preceding operations are done before all,

A. We start with the digits that are left in the given problem:

$-3+(\sqrt{100}-3^2-1^{14}):30+3$First, we simplify the digits that are in the denominators on which the division operation takes place, this is done in accordance with the order of operations mentioned, keeping in mind that multiplication precedes subtraction, therefore, we first calculate their numerical values of the numerators in the multiplication (this within the definition of the root as a power, the root of which is a power for everything), subsequently we perform the subtraction operation which is within the denominators and finally we perform the division operation that takes place on the denominators:

$-3+(\sqrt{100}-3^2-1^{14}):30+3 =\\ -3+(10-9-1):30+3 =\\ -3+0:30+3 =\\ -3+0+3 \\$In the last step we mentioned that dividing the number 0 by any number (different from zero) will yield the result 0, we continue with the simple digits we received in the last step and perform the addition operation:

$-3+0+3 =\\ 0$We finish the simple digits that are left in the given problem, we summarize the steps of the simplification:

We received that:

$-3+(\sqrt{100}-3^2-1^{14}):30+3 =\\ -3+0:30+3 =\\ 0$

B. We continue and simplify the digits that are right in the given problem:

$6^2:6\cdot(3-2)-6$In this part, in the first step we simplify the digits within the framework of the order of operations,

In this digit, multiplication takes place on digits in the denominators, therefore, we first simplify this digit, in the process we calculate the numerical values of the numerator in the multiplication which is the divisor in the first digit from the left in the given digit:

$6^2:6\cdot(3-2)-6 =\\ 36:6\cdot1-6 \\$We continue and remember that multiplication and division precede addition and subtraction, keeping in mind that there is no predefined precedence between multiplication and division operations originating from the order of operations mentioned, therefore, we calculate the numerical values of the numerator in the first digit from the left (including the multiplication and division operations) within the execution of one operation after another according to the order from left to right (this is the natural order of operations), subsequently, we complete the calculation within the execution of the subtraction operation:

$36:6\cdot1-6 =\\ 6\cdot1-6 =\\ 6-6 =\\ 0$We finish the simple digits that are right in the given problem, we summarize the steps of the simplification:

We received that:

$6^2:6\cdot(3-2)-6 =\\ 36:6\cdot1-6 =\\ 0$We return to the original problem, and we present the results of the simplifications reported in A and B:

$-3+(\sqrt{100}-3^2-1^{14}):30+3\text{ }\text{\textcolor{red}{\_\_}}\text{ }6^2:6\cdot(3-2)-6 \\ \downarrow\\ 0\text{ }\text{\textcolor{red}{\_\_}}\text{ }0$As a result, we have that:

$0 \text{ }\textcolor{red}{=}0$Therefore, the correct answer here is answer B.

According to the given problem, whether it is discussed in addition or subtraction each of the terms that comes in its turn,

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 all others,

A. We start with the terms that are on the left in the given problem:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)$First, we simplify the terms that are in the denominators (the divisors) on which the multiplication operation is performed, this in accordance with the order of operations mentioned, we note that this term will change in terms that are in the denominators (the numerators) on which division is performed, therefore we start with simply the terms that are in these denominators, remembering that division precedes multiplication and subtraction, therefore the beginning will perform the division operation that is in this term and then perform the operations of multiplication and subtraction:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big((5+1-8)^2:\sqrt{4}-2\big)=\\ \frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)\\$Since the results of the operations of multiplication and subtraction that are in the numerators the level will come out smoothly on these denominators, we continue and perform the strength on the term that is in these denominators, this within that we remember thatthe raising of any number (positive or negative) in a double strength will give a positive result, in contrast we will consider its numerical value of the other side that in strength he is the divisor that is in the term that within the denominators the divisors that were left (this within that we remember that in defining the root as strength, the root he is strength for everything):

$\frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)=\\ \frac{1}{5}\cdot\big(4:2-2\big)\\$We continue and finish simply the term, we remember that division precedes subtraction and therefore the beginning will calculate the result of the division operation that in the term and then perform the operation of subtraction:

$\frac{1}{5}\cdot\big(2-2\big)=\\ \frac{1}{5}\cdot0=\\ 0$In the last stage we performed the doubling that was left (it is the doubling that is performed on the term that in the denominators), this within that we remember thatthe result of doubling any number (different from zero) in zero here zero,

$$We finished simply the term that is on the left in the given problem, we will summarize the stages of simply:

We received that:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big(2-2\big)=\\ 0$

B. We continue and simplify the term that is on the right in the given problem:

$8-(3^2+1)\cdot\frac{1}{10}$In this part to do in the first part simplify the term within the framework of the order of operations,

In this term a doubling that is performed on the term that in the denominators, therefore we simplify first this term, this is done in accordance with the order of operations mentioned, therefore we start from considering its numerical value that in strength that in this term and then perform the operation of multiplication:

$8-(3^2+1)\cdot\frac{1}{10} =\\ 8-(9+1)\cdot\frac{1}{10}=\\ 8-10\cdot\frac{1}{10}\\$We continue and simplify the term that was received in the first stage, we remember in this that multiplication and division precede addition and subtraction, therefore the beginning will perform the doubling in the break, this within that we remember that the doubling in the break means the doubling in the amount of the break, then perform the operation of division that of the break, this is done by appointment, in the last stage will perform the operation of subtraction that remained:

$8-10\cdot\frac{1}{10}=\\ 8-\frac{10\cdot1}{10}=\\ 8-\frac{\not{10}}{\not{10}}=\\ 8-1=\\ 7$We finished simply the term that is on the right in the given problem, we will summarize the stages of simply:

We received that:

$8-(3^2+1)\cdot\frac{1}{10} =\\ 8-10\cdot\frac{1}{10}\\ 7$We return now to the original problem, and we will present the results of simply the terms that were reported in A and in B:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\text{ }_{\textcolor{red}{——}\text{\textcolor{red}{ }}}8-(3^2+1)\cdot\frac{1}{10} \\ \downarrow\\ 0\text{ }\text{\textcolor{red}{\_\_}}\text{ }7$Therefore the correct answer here is answer A.

In order to solve the given problem, whether it involves addition or subtraction each of the terms that appear in the equation must be dealt with 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 for all,

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

$-5+5-3\cdot2+6$ Simplify the terms that are in the parentheses in accordance with the order of operations, start with the multiplication that is in the terms and continue to perform the operations of addition and subtraction:

$-5+5-3\cdot2+6=\\ -5+5-6+6=\\ 0$$$

We finish simplifying the terms that are on the left in the given problem.

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

$\big((3+2)\cdot2\big):2\cdot0$Note that in this term there is a multiplication between the term and the number 0, in addition note that in this term it is defined (since it does not include division by 0), we remember that multiplying any number by 0 will yield the result 0, therefore:

$\big((3+2)\cdot2\big):2\cdot0 =\\ 0$

$$We return now to the original problem, and we will present the results of simplifying the terms that were reported in A and B:

$-5+5-3\cdot2+6\textcolor{red}{☐}\big((3+2)\cdot2\big):2\cdot0 \\ \downarrow\\ 0\text{ }\textcolor{red}{_—}0$As a result that we find that:

$0 \text{ }\textcolor{red}{=}0$Therefore the correct answer here is answer A.

We solve the left side and start from the parentheses:

$5^2=5\times5=25$

We will solve the root exercise using the equation:$\sqrt{a^2}=a$

$\sqrt{9}=\sqrt{3^2}=3$

We arrange the exercise accordingly:

$\frac{1}{25}\times(25-3+3)=$

We solve the exercise in parentheses from left to right:

$\frac{1}{25}\times(22+3)=\frac{1}{25}\times25$

We convert the 25 into a simple fraction, multiply and divide:

$\frac{1}{25}\times\frac{25}{1}=\frac{25}{25}=\frac{1}{1}=1$

We solve the right side:

$\sqrt{25}=\sqrt{5^2}$

We arrange the exercise:

$\sqrt{5^2}\times5\times\frac{1}{5}$

We convert the 5 into a simple fraction and note that it is possible to reduce by 5:

$\sqrt{5^2}\times\frac{5}{1}\times\frac{1}{5}=\sqrt{5^2}\times1$

We solve the root according to the formula:$\sqrt{a^2}=a$

$5\times1=5$

Now we are going to compare the left side with the right side, and it seems that we obtained two different results and therefore the two sides are not equal.

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:

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

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