Mastering Order of Operations: BIDMAS | 掌握运算顺序：BIDMAS

📚 Mastering Order of Operations: BIDMAS | 掌握运算顺序：BIDMAS

Welcome to the essential guide on BIDMAS – the rule that tells you which operation to perform first in a mathematical expression. Whether you are simplifying 2 + 3 × 4 or tackling a much trickier calculation with brackets and powers, BIDMAS keeps everyone on the same page. Without it, the same expression could produce a dozen different answers, and maths would descend into chaos. This article unpacks every layer of the convention, from simple brackets to combined operations with negative numbers, providing plenty of examples, common pitfalls, and practice ideas along the way.

欢迎阅读这份有关 BIDMAS 的必备指南——这条规则告诉你在一个数学表达式中应该首先进行哪种运算。无论你是在化简 2 + 3 × 4，还是在解决一个带有括号和幂的复杂计算，BIDMAS 都能确保所有人得出相同的答案。没有它，同样的表达式可能会得出十几种不同的结果，数学就会陷入混乱。本文将逐层剖析这套约定，从简单的括号一直到包含负数的混合运算，提供大量示例、常见陷阱以及练习思路。

1. What is BIDMAS? | 什么是 BIDMAS？

BIDMAS is an acronym that stands for Brackets, Indices, Division, Multiplication, Addition, Subtraction. It describes the universal order in which mathematical operations must be performed so that every mathematician, teacher, and student arrives at the same result. The letters are sometimes rearranged into BODMAS (where O stands for Orders) or PEMDAS in some countries, but the core principle never changes: higher-priority operations are done before lower-priority ones.

BIDMAS 是一个首字母缩略词，代表括号 (Brackets)、指数 (Indices)、除法 (Division)、乘法 (Multiplication)、加法 (Addition)、减法 (Subtraction)。它描述了数学运算必须遵循的通用顺序，确保每位数学家、教师和学生都能得到相同的答案。这些字母有时会被重新排列为 BODMAS（其中 O 表示 Order，即阶 / 指数），或在一些国家变成 PEMDAS，但核心原则始终不变：优先级高的运算要在优先级低的运算之前执行。

Think of BIDMAS as the grammar of arithmetic. Just as ‘The cat sat on the mat’ means something different from ‘Sat the on mat cat the’, mathematical meaning collapses if the sequence of operations is scrambled. The convention was gradually standardised from the 16th century onwards as algebraic notation developed, and today it is one of the first formal rules introduced in KS3 mathematics.

可以把 BIDMAS 想象成算术的语法。正如 ‘The cat sat on the mat’ 与 ‘Sat the on mat cat the’ 含义完全不同，如果运算顺序被打乱，数学的意义也会随之崩塌。这一约定自 16 世纪起随着代数符号的发展逐步标准化，如今已成为 KS3 数学中最早引入的正式规则之一。

2. Brackets First | 括号优先

Brackets group parts of a calculation that must be treated as a single package. They always take the highest priority: whatever is inside brackets must be worked out completely before you use that result in the surrounding expression. This includes parentheses ( ), square brackets [ ], and sometimes curly brackets { }. In KS3, you will mostly see round brackets, but the principle is identical for all types.

括号将必须视为一个整体的计算部分组合在一起。它们始终具有最高优先级：括号内的所有内容必须全部计算完毕，之后才能将其结果用于外围表达式。这包括圆括号 ( )、方括号 [ ]，有时还有花括号 { }。在 KS3 阶段，你见到的大多是圆括号，但所有类型的原则都相同。

For example, in 7 × (2 + 3), the addition 2 + 3 is done first, giving 5, and then 7 × 5 = 35. If you ignored the brackets and worked from left to right, you might wrongly do 7 × 2 = 14, then 14 + 3 = 17 – a completely different answer.

例如，在 7 × (2 + 3) 中，首先要计算加法 2 + 3 得到 5，然后再计算 7 × 5 = 35。如果你忽略括号并单纯从左往右计算，可能会错误地先算 7 × 2 = 14，然后 14 + 3 = 17——结果完全不同。

Nested brackets occasionally appear: for instance, 3 × [4 + (6 ÷ 2)]. Here you start with the innermost brackets: 6 ÷ 2 = 3, then 4 + 3 = 7, and finally 3 × 7 = 21. Always work from the inside out.

偶尔会出现嵌套括号，例如 3 × [4 + (6 ÷ 2)]。这里你要从最内层括号开始：6 ÷ 2 = 3，然后 4 + 3 = 7，最后 3 × 7 = 21。始终要从内向外计算。

3. Indices (Powers and Roots) | 指数（幂与根）

After brackets, the next priority is indices – another word for powers, exponents, or roots. Common examples include 2³ (2 × 2 × 2), 5² (5 × 5), or √9 (the square root of 9). In the BIDMAS rule, any index attached to a number or a bracket must be dealt with before moving to division or multiplication.

处理完括号之后，下一个优先级是指数——也就是幂、指数或根。常见的例子包括 2³（2 × 2 × 2）、5²（5 × 5）或 √9（9 的平方根）。按照 BIDMAS 规则，在进入除法或乘法之前，必须先处理附加在数字或括号上的任何指数。

Consider 3 + 2² × 4. The index 2² comes before either multiplication or addition, so 2² = 4. The expression becomes 3 + 4 × 4. Next, multiplication takes precedence over addition: 4 × 4 = 16, then 3 + 16 = 19. If you did the addition first, you would get 5 × 4 = 20, which would be incorrect.

考虑 3 + 2² × 4。指数 2² 先于乘法和加法，所以 2² = 4。表达式变为 3 + 4 × 4。接下来，乘法优先于加法：4 × 4 = 16，然后 3 + 16 = 19。如果先算加法，就会得到 5 × 4 = 20，这是错误的。

Roots behave exactly like powers. √(16 + 9) is different from √16 + 9. In the first, the brackets force 16 + 9 = 25, then √25 = 5. In the second, √16 = 4, then 4 + 9 = 13. This illustrates why brackets are sometimes written explicitly even when they are not strictly required – for clarity.

根号的运算方式与幂完全一样。√(16 + 9) 和 √16 + 9 截然不同。前者中，括号强制先算 16 + 9 = 25，再求 √25 = 5。后者中，√16 = 4，再加 9 得到 13。这也解释了为什么有时即使语法上并非必须，人们也会明确写出括号——为了清晰。

4. Division and Multiplication | 除法与乘法

Division and multiplication share the same priority level. Once all brackets and indices are resolved, you perform division and multiplication as they appear from left to right. A common misconception is that ‘M’ comes before ‘D’ in BIDMAS and therefore multiplication always beats division. That is not true – they are equal partners.

除法和乘法享有相同的优先级。一旦所有括号和指数都处理完毕，你就要按照从左到右的顺序进行除法和乘法运算。一个常见的误解是，BIDMAS 中 M 在 D 前面，因此乘法总是优先于除法。事实并非如此——它们是平等的。

Let’s test: 20 ÷ 4 × 5. If you mistakenly multiply first, 4 × 5 = 20, then 20 ÷ 20 = 1. The correct left-to-right approach gives 20 ÷ 4 = 5, then 5 × 5 = 25. The answers are worlds apart, so remembering ‘left to right for division and multiplication’ is crucial.

我们来测试一下：20 ÷ 4 × 5。如果错误地先乘，4 × 5 = 20，再 20 ÷ 20 = 1。而正确的从左到右做法是 20 ÷ 4 = 5，再 5 × 5 = 25。结果天差地别，因此牢记“除法和乘法从左到右”至关重要。

Why does this happen? Division can be thought of as multiplying by the reciprocal, so 20 ÷ 4 × 5 is the same as 20 × (1/4) × 5 = 25. The expression is unambiguous only when we follow the left-to-right convention. Whenever you face a string of × and ÷, simply work step by step in the reading direction.

为什么会这样呢？除法可以视为乘以倒数，所以 20 ÷ 4 × 5 等价于 20 × (1/4) × 5 = 25。只有当我们遵循从左到右的约定时，这个表达式才是明确的。每当你碰到一连串 × 和 ÷ 时，只需按阅读方向一步步计算即可。

5. Addition and Subtraction | 加法与减法

Finally, addition and subtraction are dealt with last, and again with equal priority, moving from left to right. After tidying up brackets, indices, divisions and multiplications, the remaining calculation often looks much simpler: a string of numbers joined by + and – signs.

最后，加法和减法最后处理，同样是平等优先级，从左到右依次计算。在清理完括号、指数、除法和乘法之后，剩下的计算通常看起来简单得多：一串由 + 和 − 号连接的数字。

Take 15 − 3 + 2. Working left to right, 15 − 3 = 12, then 12 + 2 = 14. If you incorrectly added first, you would get 3 + 2 = 5, then 15 − 5 = 10 – wrong. The habit of seeing ‘−’ as belonging to the number that follows helps: think of the expression as 15 + (−3) + 2, which clearly sums to 14.

以 15 − 3 + 2 为例。从左到右计算，15 − 3 = 12，然后 12 + 2 = 14。如果你错误地先做加法，就会得到 3 + 2 = 5，再 15 − 5 = 10——错了。养成“− 号归属于其后的数字”这一习惯会很有帮助：把表达式看作 15 + (−3) + 2，求和结果显然是 14。

Subtraction is simply adding a negative number, so the order of addition and subtraction can be flexible if you treat all signs correctly. However, at KS3, the safest approach is to respect the left-to-right rule until you are completely comfortable rewriting expressions as sums of positive and negative terms.

减法本质上就是加上一个负数，因此如果你正确看待所有符号，加减法的顺序其实可以灵活处理。不过在 KS3 阶段，最安全的方法还是遵守从左到右的规则，直到你完全习惯将表达式改写为正数项和负数项之和。

6. Left to Right Rule for Equal Precedence | 同级运算的左至右规则

The left-to-right rule only applies when two operations share the same rank – for instance, two multiplications, two divisions, or a mix of multiplication and division, as well as addition and subtraction. It never overrides the hierarchy: you cannot do a left-to-right addition before a right-hand multiplication.

同级运算的左至右规则仅适用于两个运算共享同一优先级的情况——例如两个乘法、两个除法，或乘除混合，以及加减混合。它绝不能推翻运算层级：你不能因为从左到右，就把左边的加法提到右边的乘法之前去算。

This idea can be practised with slightly more involved examples. Consider 30 ÷ 5 × 2 ÷ 3. Left to right: 30 ÷ 5 = 6, 6 × 2 = 12, 12 ÷ 3 = 4. Any other grouping, like 30 ÷ (5 × 2) ÷ 3, would change the meaning entirely and is only valid if brackets are physically present.

这个思路可以通过稍微复杂一点的例子来练习。考虑 30 ÷ 5 × 2 ÷ 3。从左到右：30 ÷ 5 = 6，6 × 2 = 12，12 ÷ 3 = 4。任何其他分组方式，比如 30 ÷ (5 × 2) ÷ 3，都会完全改变原意，只有在确实有括号时才有效。

A helpful tip: when you are solving a long expression, underline each step or rewrite the simplified expression below. For 48 − 12 ÷ 3 + 8 × 2, identify the higher-priority operations first: 12 ÷ 3 = 4 and 8 × 2 = 16, so the expression becomes 48 − 4 + 16. Now left-to-right addition/subtraction gives 44 + 16 = 60.

一个有用的技巧：在解长表达式时，可以下划线标出每一步，或在下方重新写出化简后的表达式。对于 48 − 12 ÷ 3 + 8 × 2，首先找出优先级较高的运算：12 ÷ 3 = 4，8 × 2 = 16，因此表达式变为 48 − 4 + 16。然后从左到右加减得到 44 + 16 = 60。

7. Common Mistakes to Avoid | 常见错误避免

Even students who know BIDMAS can slip up when operations are written in a misleading way. One typical error is treating ‘− 2²’ as (−2)². In BIDMAS, indices apply before the minus sign, so −2² means −(2²) = −4, whereas (−2)² = 4. Always check whether the base includes a negative sign or not.

即使知道 BIDMAS 的学生，当运算以误导性的方式书写时也可能出错。一个典型错误是把 −2² 当作 (−2)²。按照 BIDMAS，指数的优先级高于减号，因此 −2² 表示 −(2²) = −4，而 (−2)² = 4。一定要检查底数是否包含负号。

Another common slip-up is with fractions written as a ÷ b ÷ c. For example, 18 ÷ 3 ÷ 2 is left-to-right: 18 ÷ 3 = 6, 6 ÷ 2 = 3. Some learners mistakenly think it’s 18 ÷ (3 ÷ 2) = 18 ÷ 1.5 = 12. The absence of brackets makes the first interpretation correct.

另一个常见错误与写成分数形式的 a ÷ b ÷ c 有关。例如，18 ÷ 3 ÷ 2 应从左到右：18 ÷ 3 = 6，6 ÷ 2 = 3。有些学习者却错误地认为它是 18 ÷ (3 ÷ 2) = 18 ÷ 1.5 = 12。由于没有括号，第一种理解才是正确的。

Also watch out for hidden brackets in fraction bars or roots that stretch over a long expression. A large square root symbol √(25 + 144) has an invisible bracket to group 25 + 144, so you must add first. Similarly, a fraction like (2 + 4)/(3 − 1) implies brackets around the numerator and denominator.

还要留意分数线和覆盖较多内容的根号所暗含的括号。一个大的平方根符号 √(25 + 144) 实际上有一个隐含括号把 25 + 144 括在一起，因此必须先做加法。类似地，像 (2 + 4)/(3 − 1) 这样的分数意味着分子和分母各有隐含括号。

8. Complex Examples Step by Step | 复杂例题逐步解析

Let’s walk through a few examination-style problems that draw on the full range of BIDMAS. These will give you a clear template for setting out your work.

我们来逐步讲解几道覆盖全部 BIDMAS 等级的考试型题目。这将为你提供清晰的书写范本。

Example 1: 5 × (2 + 3)² − 6 ÷ 2
Step 1 – Brackets: (2 + 3) = 5.
Step 2 – Indices: 5² = 25.
Step 3 – Multiplication and Division: 5 × 25 = 125, 6 ÷ 2 = 3.
Step 4 – Subtraction: 125 − 3 = 122.
Correct answer: 122.

示例 1：5 × (2 + 3)² − 6 ÷ 2
第 1 步——括号：(2 + 3) = 5。
第 2 步——指数：5² = 25。
第 3 步——乘除：5 × 25 = 125，6 ÷ 2 = 3。
第 4 步——减法：125 − 3 = 122。
正确答案：122。

Example 2: 40 − 3 × (8 − 2)² ÷ 9
Step 1 – Inner bracket: 8 − 2 = 6.
Step 2 – Index: 6² = 36.
Step 3 – Multiplication and Division left to right: 3 × 36 = 108, then 108 ÷ 9 = 12.
Step 4 – Subtraction: 40 − 12 = 28.
Correct answer: 28.

示例 2：40 − 3 × (8 − 2)² ÷ 9
第 1 步——内层括号：8 − 2 = 6。
第 2 步——指数：6² = 36。
第 3 步——从左到右乘除：3 × 36 = 108，然后 108 ÷ 9 = 12。
第 4 步——减法：40 − 12 = 28。
正确答案：28。

Example 3: (12 + 4)² ÷ √(64) − 5 × 2 + 10
Step 1 – Brackets: 12 + 4 = 16. Also, inside the root, 64 is already alone, so no extra bracket work.
Step 2 – Indices/roots: 16² = 256, √64 = 8.
Step 3 – Division/Multiplication: 256 ÷ 8 = 32, 5 × 2 = 10.
Step 4 – Addition/Subtraction: 32 − 10 + 10 = 32. (The −10 and +10 cancel).
Correct answer: 32.

示例 3：(12 + 4)² ÷ √(64) − 5 × 2 + 10
第 1 步——括号：12 + 4 = 16。根号内 64 已独立，无需额外括号处理。
第 2 步——指数/根：16² = 256，√64 = 8。
第 3 步——除/乘：256 ÷ 8 = 32，5 × 2 = 10。
第 4 步——加减：32 − 10 + 10 = 32。（−10 和 +10 抵消。）
正确答案：32。

9. Using BIDMAS with Negative Numbers | BIDMAS 与负数的运用

Negative numbers add an extra layer of care, but BIDMAS still holds. The key is to treat the minus sign before a number as part of that number when you reach addition/subtraction, but be alert to whether it is an operation or a sign. For instance, in 5 − (−3), the brackets tell you to deal with −3 first, which is just −3, then subtract that: 5 − (−3) = 5 + 3 = 8.

负数增加了一层额外的注意事项，但 BIDMAS 依然成立。关键在于，在做加减时要把数字前的减号看作该数字的一部分，但同时要警惕它到底是运算符还是正负号。例如，5 − (−3) 中，括号告诉你先处理 −3，就是 −3，然后再减去它：5 − (−3) = 5 + 3 = 8。

Consider 10 + (−4) × 3. The multiplication comes before addition: (−4) × 3 = −12, then 10 + (−12) = −2. If you added first, 10 + (−4) = 6, then 6 × 3 = 18, which is wrong. BIDMAS always directs you to multiply before you add, irrespective of negative signs.

考虑 10 + (−4) × 3。乘法优先于加法：(−4) × 3 = −12，然后 10 + (−12) = −2。如果先做加法，10 + (−4) = 6，再 6 × 3 = 18，就错了。BIDMAS 始终要求先乘后加，与负数符号无关。

Another delicate situation is −2² as mentioned earlier. Without brackets, the index applies only to the 2, not to the minus sign. So −2² = −4. Always place brackets around the base when you intend the negative to be squared: (−2)² = 4.

另一种棘手情况是前面提到的 −2²。没有括号时，指数仅作用于 2，而不作用于负号。因此 −2² = −4。如果你希望将负号一同平方，一定要用括号标出底数：(−2)² = 4。

10. Practice Questions | 练习题

To cement your understanding, try these BIDMAS challenges. Calculate each step carefully and check your answers against the solutions below.

为了巩固理解，请尝试以下 BIDMAS 挑战。仔细计算每一步，并对照下面的答案进行核对。

1. 7 + 3 × (10 − 6) ÷ 2
2. (8 − 3)² + 14 ÷ 7 − 1
3. 5² − 4 × 3 + 12 ÷ 2²
4. 20 − (6 + 2 × 3) + √9
5. −3 × (−4) + 10 ÷ (−2) + 6

Solutions:
1. 7 + 3 × 4 ÷ 2 = 7 + 12 ÷ 2 = 7 + 6 = 13.
2. 5² + 2 − 1 = 25 + 2 − 1 = 26.
3. 25 − 12 + 12 ÷ 4 = 25 − 12 + 3 = 16.
4. 20 − (6 + 6) + 3 = 20 − 12 + 3 = 11.
5. 12 + (−5) + 6 = 13.

答案：
1. 7 + 3 × 4 ÷ 2 = 7 + 12 ÷ 2 = 7 + 6 = 13。
2. 5² + 2 − 1 = 25 + 2 − 1 = 26。
3. 25 − 12 + 12 ÷ 4 = 25 − 12 + 3 = 16。
4. 20 − (6 + 6) + 3 = 20 − 12 + 3 = 11。
5. 12 + (−5) + 6 = 13。

11. Real-Life Applications of Order of Operations | 运算顺序的现实应用

It might feel as though BIDMAS only lives inside maths classrooms, but the order of operations appears whenever a single formula combines several different actions. In finance, compound interest calculations mix powers and multiplications: A = P(1 + r/n)^(nt). The exponent must be evaluated first, then the contents of the bracket, then the multiplication – exactly as BIDMAS dictates.

或许你会觉得 BIDMAS 只存在于数学课堂，但只要一个公式结合了多种不同的运算，运算顺序就会出现。在金融中，复利计算把幂和乘法交织在一起：A = P(1 + r/n)^(nt)。必须先计算指数，再处理括号内的内容，然后进行乘法——完全遵循 BIDMAS。

Computer programming languages also rely on operator precedence rules that mirror BIDMAS. When you code a line like distance = initial velocity * time + 0.5 * acceleration * time ** 2, the interpreter squares the time first (indices), multiplies, and then adds – exactly the same logic.

计算机编程语言也依赖于反映 BIDMAS 思想的运算符优先级规则。当你编写类似 distance = initial velocity * time + 0.5 * acceleration * time ** 2 的代码时，解释器会先对时间求平方（指数），再做乘法，最后相加——逻辑完全相同。

Even everyday spreadsheets obey these rules. Typing =2+3*4 into a spreadsheet cell gives 14, not 20, because the software is programmed to follow standard mathematical conventions. Understanding BIDMAS therefore helps you both with exams and with tools you will use for life.

就连日常电子表格也遵循这些规则。在单元格中输入 =2+3*4 会得到 14，而不是 20，因为软件已按标准数学约定编程。理解 BIDMAS 因此既能帮你应对考试，也能助你运用终身的工具。

12. Summary and Key Takeaways | 总结与要点

BIDMAS is the backbone of calculation. The hierarchy is fixed: Brackets → Indices → Division & Multiplication (left to right) → Addition & Subtraction (left to right). Every problem, no matter how complicated, can be broken down step by step by applying these rules in the correct order.

BIDMAS 是计算的支柱。层级是固定的：括号 → 指数 → 除法和乘法（从左到右）→ 加法和减法（从左到右）。无论问题多么复杂，都可以通过按照正确顺序应用这些规则，一步步地分解解决。

Key points to remember:

Brackets always come first – resolve everything inside them.
Indices (powers and roots) are next.
Division and multiplication are equal – work left to right.
Addition and subtraction are equal – work left to right.
Treat a minus sign as belonging to the number that follows.
Use step-by-step rewriting to avoid mistakes on long expressions.

关键要点：

括号始终优先——计算括号内的全部内容。
指数（幂和根号）紧随其后。
除法与乘法同级——从左到右计算。
加法与减法同级——从左到右计算。
把减号视为跟随其后的数字的一部分。
面对长表达式，用逐步改写的方式避免错误。

With regular practice, BIDMAS will become second nature, transforming confusing strings of symbols into manageable, logical steps that always lead to the correct answer.

通过经常练习，BIDMAS 将成为你的第二天性，把令人困惑的符号串转变为易于管理、合乎逻辑的步骤，并始终导向正确答案。

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Mastering Order of Operations: BIDMAS | 掌握运算顺序：BIDMAS