Special Cases (0 and 1, Inverse, Fraction Line): Parentheses within parentheses

In the order of operations, parentheses come before everything else.

We start by solving the inner parentheses in the subtraction operation:

$((3):3-1)\times4=$ We continue with the inner parentheses in the division operation and then subtraction:

$(1-1)\times4=$

We continue solving the subtraction exercise within parentheses and then multiply:

$0\times4=0$

Simplifying this expression emphasizes the order of operations, which states that multiplication precedes addition and subtraction, and that division precedes all of them,

In the given expression, the establishment of division operations between the parentheses (the outermost) to a number, therefore according to the order of operations as mentioned, is handled by simplifying the expression in these parentheses, this expression includes division operations that begin on the expression within the parentheses (the innermost), therefore according to the order of operations as mentioned is handled by simplifying the expression in these parentheses and performing the subtraction operations in it, there is no hindrance to calculate the outcome of the division operations in the expression in the outermost parentheses, but for the sake of good order this is done afterwards:

$\lbrack(4+3):7+2:2-2\rbrack:5= \\ \lbrack7:7+2:2-2\rbrack:5$Continuing and simplifying the expression in the parentheses we noted, since division precedes addition and subtraction, start with the division operations in the expression and only then calculate the outcome of the addition and subtraction, ultimately perform the division operations on this expression in the parentheses:

$\lbrack7:7+2:2-2\rbrack:5 \\ \lbrack1+1-2\rbrack:5=\\ 0:5=\\0$In the last stage we mentioned that multiplying a number by 0 gives the result 0,

Therefore, this simplifying expression is short so there is no need to elaborate,

And the correct answer here is answer A.

This simple rule is the order of operations which states that multiplication and division come before addition and subtraction, and operations enclosed in parentheses come first,

In the given example of division between two given numbers in parentheses, therefore according to the order of operations mentioned above, we start by calculating the values of each of the numbers within the parentheses, there is no prohibition against calculating the result of the addition operation in the given number, for the sake of proper order, this operation is performed later:

$(3+2-1):(1+3)-1+5= \\ 4:4-1+5$In continuation of the principle that division comes before addition and subtraction the division operation is performed first and then the operations of subtraction and addition which were received in the given number and in the last stage:

$4:4-1+5= \\ 1-1+5=\\ 5$Therefore the correct answer here is answer B.

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.

This simple example demonstrates the order of operations, which states that exponentiation precedes multiplication and division, which come before addition and subtraction, and that operations within parentheses come first,

In the given example, the operation of division between parentheses (the denominators) by a number (which is also in parentheses but only for clarification purposes), thus according to the order of operations mentioned we start with the parentheses that contain the denominators first, this parentheses that contain the denominators includes multiplication between two numbers which are also in parentheses, therefore according to the order of operations mentioned, we start with the numbers inside them, paying attention that each of these numbers, including the ones in strength, and therefore assuming that exponentiation precedes multiplication and division we consider their numerical values only in the first step and only then do we perform the operations of multiplication and division on these numbers:

$$$$$\lbrack(3^2-4-5)\cdot(4+\sqrt{16})-5 \rbrack:(-5)=\\ \lbrack(9-4-5)\cdot(4+4)-5 \rbrack:(-5)=\\ \lbrack0\cdot8-5 \rbrack:(-5)\\$Continuing with the simple division in parentheses ,and according to the order of operations mentioned, we proceed from the multiplication calculation and remember that the multiplication of the number 0 by any number will yield the result 0, in the first step the operation of subtraction is performed and finally the operation of division is initiated on the number in parentheses:

$\lbrack0\cdot8-5 \rbrack:(-5)= \\ \lbrack0-5 \rbrack:(-5)= \\ -5 :(-5)=\\ 1$ Therefore, the correct answer is answer c.

According to the rules of order of arithmetic operations, parentheses are resolved first.

We start by solving the inner parentheses, first we will solve the root using the formula:

$\sqrt{a}=\sqrt{a^2}=a$

$\sqrt{81}=\sqrt{9^2}=9$

The exercise obtained within parentheses is:

$(9-3\times3)$

First we solve the multiplication exercise and then we subtract:

$(9-9)=0$

After solving the inner parentheses, the resulting exercise is:

$0:4+5\times5$

According to the rules of the order of arithmetic operations, we first solve the exercises of multiplication and division, and then subtraction.

We place the two exercises within parentheses to avoid confusion:

$(0:4)+(5\times5)=0+25=25$

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.

Let's solve the given expression step-by-step, applying the order of operations, commonly known by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).
The given expression is:
$\frac{[(25+3\cdot2)-6]:5}{4+1}-\frac{76}{19}$

Step 1: Solve inside the innermost parentheses
We start by solving the expression inside the square brackets $[\cdot]$:

• Simplify: $3\cdot2 = 6$
• Add: $25 + 6 = 31$
• Subtract: $31 - 6 = 25$

Step 2: Solve the division inside the fraction
Next, we calculate:

• Divide: $25:5 = 5$

Step 3: Solve the denominator of the fraction

• Add: $4 + 1 = 5$

Step 4: Solve the subtraction
Finally, perform the subtraction:

• Subtract: $1 - \frac{76}{19}$

• Subtract: $1 - 4 = -3$

The solution to the question is: -3

This simple rule is the emphasis on the order of operations which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others,

Let's consider that the numerator is the whole and the denominator is the part which breaks (every break) into whole pieces (in their entirety) among which division operation is performed, meaning- we can relate the numerator and the denominator of the break as whole pieces in closures, thus we can express the given fraction and write it in the following form:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}= \\ \downarrow\\ \big((25-2-16)^2+3\big):(8+5):\sqrt{9}$We highlight this by noting that fractions in the numerator of the break and in its denominator are considered separately, as if they are in closures,

Let's return to the original fraction in question, meaning - in the given form, and simplify, separately, the fraction in the numerator of the break which causes it and the fraction in its denominator, this is done in accordance to the order of operations mentioned and in a systematic way,

Let's consider that in the numerator of the break the fraction we get changes into a fraction in closures which indicates strength, therefore we will start simplifying this fraction, given that this fraction includes only addition and subtraction operations, perform the operations in accordance to the natural order of operations, meaning- from left to right, simplifying the fraction in the numerator of the break:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}=\\ \frac{7^2+3}{13}:\sqrt{9}\\$We will continue and simplify the fraction we received in the previous step, this of course, in accordance to the natural order of operations (which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others), therefore we will start from calculating the numerical values of the exponents in strength (while we remember that in defining the root as strength, the root itself is strength for everything), and then perform the division operation which is in the numerator of the break:

$\frac{7^2+3}{13}:\sqrt{9}=\\ \frac{49+3}{13}:3=\\ \frac{52}{13}:3\\$We will continue and simplify the fraction we received in the previous step, starting with performing the division operation of the break, this is done by approximation, and then perform the remaining division operation:

$\frac{\not{52}}{\not{13}}:3=\\ 4:3=\\ \frac{4}{3}$In the previous step, given that the outcome of the division operation is different from a whole (greater than whole for the numerator, given that the divisor is greater than the dividend) we marked its outcome as a fraction in approximation (where the numerator is greater than the denominator),

We conclude the steps of simplifying the given fraction, we found that:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}=\\ \frac{7^2+3}{13}:\sqrt{9}=\\ \frac{52}{13}:3=\\ \frac{4}{3}$Therefore, the correct answer is answer b'.

Note:

Let's consider that in the group of the previous steps in solving the problem, we can start recording the break and the division operation that affects it even without the break, but with the help of the division operation:

$\frac{52}{13}:3\\ \downarrow\\ 52:13:3$And from here on we will start calculating the division operation in the break and only after that we performed the division in number 3, we emphasize that in general we simplify this fraction in accordance to the natural order of operations, meaning we perform the operations one after the other from left to right, and this means that there is no precedence of one division operation in the given fraction over the other except as defined by the natural order of operations, meaning- in calculating from left to right, (Let's consider additionally that defining the order of operations mentioned at the beginning of the solution, which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others, does not define precedence even among multiplication and division, and therefore the judgment between these two operations, in different closures, is in a different order, it is in calculating from left to right).

This simple example demonstrates the order of operations, which states that multiplication and division take precedence over addition and subtraction, and that operations within parentheses take precedence over all others,

Let's consider the numerator and the denominator separately (each separately) which between them performs a division operation, meaning- we can relate to the numerator and the denominator separately as fractions in their own right, thus we can write the given fraction and write it in the following form:

$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \downarrow\\ \big((25-2-16)^2+3\big):\big((\sqrt{9}\cdot7):\sqrt{3} \big)$We emphasize this by noting that the fractions in the numerator and the denominator should be treated separately, indeed as if they are in their own parentheses,

Let's consider additionally that the division operation between the parentheses implies that we are dividing by the value of the denominator (meaning the denominator as a whole, it is the result of the division between the numerator and the denominator) and therefore in the given fraction to form a division that we marked for attention, the denominator being in parentheses is additionally important,

Returning to the original fraction problem, meaning - in the given form, and proceed simply,

We will start and simplify the fraction in the numerator (meaning- the numerator fraction that we are dividing by), this is done in accordance with the order of operations mentioned above, therefore we will start by calculating the numerical values of the fraction that takes precedence (this within the context of setting the root as a priority, the root being strong for everything) and then proceed with the multiplication which is in the numerator, in contrast let's consider within the parentheses that are left, those parts in the denominator are divided by the whole, they are fractions in the stronger parentheses, therefore we will also simplify this fraction, this in accordance with the order of operations mentioned above:

$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{3\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{21}{3}\\$We will continue and simplify the fraction we received in the previous step, continue simply the fraction found within the parentheses that are divided by the whole, they are the parentheses that are left, remembering that multiplication takes precedence over addition and subtraction, therefore we will start by calculating their numerical values that take precedence in those parentheses and then proceed with the subtraction operation, in the next step the division operation of the whole (and not the division operation in the whole) takes place, and in the last step the remaining division operation takes place:

$\big(5^2-2^2\big):\frac{21}{3}=\\ \big(25-4\big):\frac{21}{3}=\\ 21:\frac{21}{3}=\\ 21:\frac{\not{21}}{\not{3}}=\\ 21:7=\\ 3$Let's consider that we advanced the division operation of the whole over the division operation in the whole itself, and this means that the number 21 in the fraction we discussed is divided by its numerical values of the whole (in its entirety)- which is the result of the division of the numerator by the denominator, therefore it was necessary to complete first the calculation of the numerical values of the whole and only then to divide the number 21 in this value,

We will conclude thus with the steps of simplifying the given fraction:
$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{21}{3}=\\ 21:7=\\ 3$Therefore, the correct answer is answer d'.