📚 Essential Maths Book 9F: Year 9 Mathematics Core Topics and Practice | 《Essential Maths 9F》Year 9 数学核心考点与练习精讲
Welcome to this Essential Maths Book 9F revision guide. It brings together the key skills that Year 9 students need to master: number work, fractions and percentages, ratio, algebra, sequences, graphs, geometry, statistics and probability. Each section explains the method, shows a worked example and gives you the phrase you need for the exam. Use it for quick review, homework support and end-of-topic tests.
欢迎使用本《Essential Maths 9F》复习指南。它汇总了 Year 9 学生必须掌握的核心技能：数与运算、分数与百分数、比、代数、数列、图像、几何、统计与概率。每一节都讲解方法、给出例题，并提炼考试中真正用得上的表述。可用于快速复习、作业辅导和单元测验备考。
Every section below pairs an English explanation with a Chinese explanation of exactly the same idea, so you can revise in either language or both at once. The worked numbers stay the same in both versions, which makes it easy to check that you have understood the reasoning rather than just the words.
下文每一节都把同一个知识点的英文讲解与中文讲解配对呈现，你可以只用一种语言复习，也可以两种对照着看。两种版本中的例题数字完全一致，便于你检验自己掌握的是解题思路，而不只是字面翻译。
1. Number and Place Value | 数系与位值
< 10 and n is a positive integer. Very small numbers use negative indices in exactly the same structure. You must also be able to round to a given number of decimal places or significant figures, and use estimation to check answers.
Year 9 在之前的基础上把位值向两端扩展。极大的数用标准形式 A × 10ⁿ 表示，其中 1 ≤ A < 10，n 为正整数；极小的数使用完全相同的结构，只是指数为负。你还需要会将数精确到指定的小数位或有效数字，并利用估算来检查答案。
4 500 000 = 4.5 × 10⁶ 0.00042 = 4.2 × 10⁻⁴
s 3.28 because the third decimal digit is 6. Significant figures count from the first non-zero digit, so 0.00476 rounded to 2 significant figures is 0.0048. Estimating 48.7 × 9.2 with 50 × 9 = 450 tells you the exact answer must be close to 450.
例如：4 500 000 写成标准形式为 4.5 × 10⁶；0.00042 写成标准形式为 4.2 × 10⁻⁴。将 3.276 精确到两位小数得 3.28，因为第三位小数是 6。有效数字从第一个非零数字算起，所以 0.00476 精确到两位有效数字为 0.0048。用 50 × 9 = 450 估算 48.7 × 9.2，说明精确答案一定接近 450。
rs follow the same rules, but be careful with signs: (−3)² = 9 while −3² = −9, because the index applies only to the 3 in the second case.
运算顺序同样重要：先括号，再乘方，然后乘除，最后加减。负数的规则相同，但要特别小心符号：(−3)² = 9，而 −3² = −9，因为第二种情况下乘方只作用于 3。
- Standard form is used in science for very large and very small quantities. 标准形式常用于科学中表示极大或极小的量。
- Estimate first, then calculate, then compare — this catches most calculator slips. 先估算、再计算、最后比较，可以抓住大多数计算器输入错误。
- Write money to 2 decimal places and always state the unit. 金额保留两位小数，并始终写出单位。
2. Fractions, Decimals and Percentages | 分数、小数与百分数
Converting between fractions, decimals and percentages is essential. You also need to find a percentage of an amount, increase or decrease by a percentage, and solve reverse percentage problems. Percentages are simply fractions with denominator 100, so every percentage can be written as a decimal multiplier.
分数、小数和百分数之间的转换至关重要。你还需要求一个数的百分之几、按百分比增加或减少，并解决逆向百分数问题。百分数本质上就是分母为 100 的分数，因此每个百分数都可以写成一个十进制乘数。
| Fraction | Decimal | Percentage |
|---|---|---|
| 1/2 | 0.5 | 50% |
| 3/4 | 0.75 | 75% |
| 1/3 | 0.333... | 33⅓% |
For percentage change, use the formula below. For example, a price rising from £40 to £48 is an increase of 20%, because the change is £8 and 8 ÷ 40 = 0.2.
计算百分比变化时使用下面的公式。例如，价格从 £40 上涨到 £48，涨幅为 20%，因为变化量是 £8，而 8 ÷ 40 = 0.2。
percentage change = (change ÷ original amount) × 100%
Reverse percentage problems work backwards. If a coat is reduced by 20% and now costs £48, the original price was £48 ÷ 0.8 = £ ify whether you are working forward or backward before calculating, and remember that a 20% reduction means the new price is 80% of the old one, not 20% less of the new price.
逆向百分数问题需要倒推。如果一件外套降价 20% 后现价为 £48，那么原价是 £48 ÷ 0.8 = £60。计算前一定要判断自己是正向还是逆向求解，并牢记：降价 20% 意味着现价是原价的 80%，而不是"再减掉现价的 20%"。
- Increase by 15%: multiply by 1.15. 增加 15%：乘以 1.15。
- Decrease by 15%: multiply by 0.85. 减少 15%：乘以 0.85。
- Find a fraction of an amount: divide by the denominator, multiply by the numerator. 求一个量的几分之几：先除以分母，再乘以分子。
- Add or subtract fractions: use a common denominator first. 分数加减：先通分。
3. Ratio, Proportion and Scale | 比、比例与比例尺
A ratio compares quantities of the same kind, and the units cancel. Simplify a ratio by dividing every part by the highest common factor, and remember that a ratio written as 3 : 5 is not the same statement as the fraction 3/5 unless you say what the whole is.
比用于比较同类量，单位会相互抵消。化简比时，把每一项都除以最大公因数。要记住，写成 3 : 5 的比与分数 3/5 并不等价，除非你说明"整体"指的是什么。
n amount in a given ratio, first add the parts to find the total number of shares. Sharing £240 in the ratio 3 : 5 gives 3 + 5 = 8 shares, so one share is £240 ÷ 8 = £30, and the two amounts are £90 and £150.
按给定比分配一个量时，先把各项相加求出总份数。将 £240 按 3 : 5 分配，共 3 + 5 = 8 份，每份为 £240 ÷ 8 = £30，所以两部分分别是 £90 和 £150。
Scale drawings use the same idea. A map scale of 1 : 50 000 means 1 cm on the map represents 50 000 cm, which is 500 m or 0.5 km in real life. Direct proportion problems are often solved by the unitary method: find the value of one item, then multiply.
比例尺图运用的是同一思路。地图比例尺 1 : 50 000 表示图上 1 cm 代表实际 50 000 cm，即 500 m 或 0.5 km。正比例问题通常用"归一法"解决：先求出一份的值，再乘上所需份数。
- Best-buy questions: compare the cost of the same unit in each offer. 性价比比较题：把每种方案都换算成同样单位的单价再比较。
- Recipe and mixture problems: keep every ingredient in the same ratio. 配方与混合问题：所有配料保持相同的比。
- Currency conversion: multiply by the exchange rate when going from pounds to another currency, divide when coming back. 货币换算：由英镑换成外币用乘法，换回英镑用除法。
4. Algebraic Expressions and Indices | 代数式与指数
Algebra in Year 9 is mostly about tidying up expressions and recognising structure. Collect like terms, multiply out brackets carefully, and factorise by taking out the highest common factor. When you expand two brackets, every term in the first bracket multiplies every term in the second.
Year 9 的代数主要是整理表达式与识别结构。合并同类项、仔细去括号，并提取最大公因式进行因式分解。展开两个括号相乘时，第一个括号中的每一项都要与第二个括号中的每一项相乘。
(x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15
re 3 and 4, so x² + 7x + 12 = (x + 3)(x + 4). Always expand mentally to check your answer.
因式分解是展开的逆过程。分解 x² + 7x + 12 时，寻找两个数，使它们的积为 12、和为 7，这两个数是 3 和 4，所以 x² + 7x + 12 = (x + 3)(x + 4)。一定要在脑中展开验算一次。
The index laws must be learned precisely, because they appear in later topics such as standard form and sequences.
指数运算法则必须精确掌握，因为它们在标准形式、数列等后续主题中都会用到。
bsp; aᵐ ÷ aⁿ = aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ a⁰ = 1 a⁻ⁿ = 1 ÷ aⁿ
- Multiply out and simplify: 3(2x − 4) + 5(x + 1) = 6x − 12 + 5x + 5 = 11x − 7. 去括号并化简：3(2x − 4) + 5(x + 1) = 11x − 7。
- Substitution: if x = −2, then x² + 3x = 4 − 6 = −2. 代入求值：若 x = −2，则 x² + 3x = 4 − 6 = −2。
- Common factor: 6x² + 9x = 3x(2x + 3). 提取公因式：6x² + 9x = 3x(2x + 3)。
5. Equations, Formulae and Inequalities | 方程、公式与不等式
Solving an equation means finding the value of the unknown that makes both sides equal. Do the same operation to both sides, and collect the unknown terms on the side that keeps the numbers positive wherever possible.
解方程就是求出使两边相等的未知数的值。对两边做相同的运算，并尽量把含未知数的项集中到能让数字保持为正的一侧。
5x − 7 = 3x + 9 → 2x = 16 → x = 8
t x = 8. Check by substituting: 5(8) − 7 = 33 and 3(8) + 9 = 33, so the answer is correct.
两边同时减去 3x 得 2x − 7 = 9，两边同时加 7 得 2x = 16，再除以 2 得 x = 8。代入验算：5(8) − 7 = 33，3(8) + 9 = 33，说明答案正确。
h sides by 2π. Inequalities are solved in the same way, with one extra rule: if you multiply or divide both sides by a negative number, the inequality sign reverses.
公式变形使用相同的步骤。
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