📚 Essential Maths 8H Homework Book Knowledge Points | KS3 数学 8H 作业本知识点精讲
The Essential Maths 8H Homework Book is designed for Year 8 pupils following a higher-tier curriculum. It covers a broad range of topics from number properties and fractions to algebra, geometry, and statistics. This article provides a structured walkthrough of the key knowledge points you will meet in this book, with clear explanations and examples to support both classroom learning and independent revision.
Essential Maths 8H 作业本是为 8 年级高阶课程学生设计的。它涵盖了从数的性质和分数到代数、几何和统计的广泛主题。本文对书中涉及的关键知识点进行了系统梳理，通过清晰的解释和例子帮助你在课堂学习和自主复习中打下扎实的基础。
1. Integers, Powers and Roots | 整数、幂与方根
Working confidently with negative numbers is a core skill at this level. When adding a negative number, you move left on the number line; subtracting a negative is the same as adding the positive. For multiplication and division, remember that two same signs give a positive result, while two different signs give a negative result.
熟练处理负数是这个阶段的核心技能。加上一个负数相当于在数轴上向左移动；减去一个负数相当于加上正数。对于乘法和除法，记住同号得正，异号得负。
Square numbers and cube numbers appear frequently. A square number is obtained by multiplying an integer by itself, e.g. 4² = 4 × 4 = 16. A cube number is obtained by multiplying an integer by itself twice, e.g. 3³ = 3 × 3 × 3 = 27. The reverse operations are square root and cube root. For instance, √64 = 8 because 8² = 64, and ∛125 = 5 because 5³ = 125.
平方数和立方数经常出现。平方数是将一个整数与自身相乘得到的，如 4² = 4 × 4 = 16。立方数是将一个整数与自身相乘两次，如 3³ = 3 × 3 × 3 = 27。逆运算分别是平方根和立方根。例如 √64 = 8，因为 8² = 64；∛125 = 5，因为 5³ = 125。
A solid understanding of prime factors is essential. Every composite number can be written uniquely as a product of prime numbers. For example, 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5. This knowledge helps with finding highest common factors (HCF) and lowest common multiples (LCM).
理解质因数是必要的。每个合数都可以唯一地写成质数的乘积。例如 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5。这一知识有助于求最大公因数（HCF）和最小公倍数（LCM）。
2. Fractions, Decimals and Percentages | 分数、小数与百分比
Equivalence between fractions, decimals and percentages is tested extensively. A fraction such as 3/5 can be converted to a decimal by dividing the numerator by the denominator (3 ÷ 5 = 0.6) and then into a percentage by multiplying by 100 (0.6 × 100 = 60%). You should memorise common equivalents such as 1/4 = 0.25 = 25% and 1/3 ≈ 0.333… = 33⅓%.
分数、小数和百分比之间的等价关系考查很广。像 3/5 这样的分数可以通过用分子除以分母（3 ÷ 5 = 0.6）转换为小数，然后乘以 100 得到百分比（0.6 × 100 = 60%）。你应该记住常见的等价关系，如 1/4 = 0.25 = 25% 和 1/3 ≈ 0.333… = 33⅓%。
Adding and subtracting fractions requires a common denominator. For mixed numbers, convert them to improper fractions first. Multiplying fractions is straightforward: multiply the numerators together and the denominators together. Dividing by a fraction means multiplying by its reciprocal. For example, 3/4 ÷ 2/5 = 3/4 × 5/2 = 15/8 = 1 7/8.
分数加减需要公分母。对于带分数，先转换为假分数。分数乘法很简单：分子相乘，分母相乘。除以一个分数等于乘以它的倒数。例如 3/4 ÷ 2/5 = 3/4 × 5/2 = 15/8 = 1 7/8。
Finding a percentage of an amount without a calculator is often done by building up from 10% or 1%. For example, to find 35% of £280, calculate 10% = £28, then 5% = £14. So 35% = 3 × £28 + £14 = £84 + £14 = £98. Percentage increase and decrease problems are also common: a 15% increase on £200 gives £200 × 1.15 = £230.
不用计算器求一个数的百分比通常从 10% 或 1% 逐步推算。例如，求 £280 的 35%，先算 10% = £28，再算 5% = £14。那么 35% = 3 × £28 + £14 = £84 + £14 = £98。百分比增减问题也很常见：£200 增加 15% 得 £200 × 1.15 = £230。
3. Ratio and Proportion | 比与比例
Ratios compare quantities in the same unit. The ratio 3:5 means for every 3 parts of one quantity there are 5 parts of another. Simplifying ratios works just like simplifying fractions — divide both sides by their highest common factor. For instance, 24:36 simplifies to 2:3 by dividing by 12.
比用来比较同单位的两个量。比例 3:5 意味着每 3 份的第一个量对应 5 份的第二个量。化简比就像化简分数一样——用最大公因数去除两边。例如 24:36 除以 12 化简为 2:3。
Sharing an amount in a given ratio involves finding the value of one part. To share £480 in the ratio 3:5, the total number of parts is 3 + 5 = 8. One part = £480 ÷ 8 = £60, so the shares are 3 × £60 = £180 and 5 × £60 = £300.
按给定比例分配金额需要先求出一份的值。将 £480 按 3:5 分配，总份数为 3 + 5 = 8。一份 = £480 ÷ 8 = £60，因此分配额为 3 × £60 = £180 和 5 × £60 = £300。
Direct proportion problems often involve scaling recipes or costs. If 5 pens cost £3.50, then the cost of 8 pens is found by working out the unit cost: £3.50 ÷ 5 = £0.70 per pen, then 8 × £0.70 = £5.60. The unitary method is a powerful tool here.
正比例问题经常涉及配方或成本的缩放。如果 5 支笔花费 £3.50，那么 8 支笔的费用可以通过求单价得出：£3.50 ÷ 5 = 每支 £0.70，然后 8 × £0.70 = £5.60。单件法是解决这类问题的有力工具。
4. Algebraic Expressions and Simplification | 代数表达式与化简
Algebra in Year 8 becomes more formal. Terms are the building blocks of an expression, separated by + or – signs. Like terms contain exactly the same letter combinations and can be collected together. For example, 3a + 5b – a + 2b simplifies to 2a + 7b.
8 年级的代数更加规范。项是表达式的基本组成，由加号或减号分隔。同类项包含完全相同的字母组合，可以合并。例如 3a + 5b – a + 2b 化简为 2a + 7b。
Multiplying terms follows index rules. a³ × a⁴ = a⁷ because you add the powers when the base is the same. b × b² = b³. When terms have coefficients, multiply numbers and letters separately: 4x² × 3x³ = 12x⁵. Expanding a single bracket uses the distributive law: 3(2x – 5) = 6x – 15.
项的乘法遵循指数法则。a³ × a⁴ = a⁷，因为同底数幂相乘指数相加。b × b² = b³。当项带有系数时，数字和字母分别相乘：4x² × 3x³ = 12x⁵。展开单项括号使用分配律：3(2x – 5) = 6x – 15。
Factorising is the reverse of expanding. Look for the highest common factor of the terms. For 10x + 15, the HCF is 5, so we write 5(2x + 3). For 8m² – 4m, the HCF is 4m, giving 4m(2m – 1).
因式分解是展开的逆运算。找出各项的最大公因式。对于 10x + 15，最大公因式是 5，因此写成 5(2x + 3)。对于 8m² – 4m，最大公因式是 4m，得到 4m(2m – 1)。
5. Solving Linear Equations | 解线性方程
Solving equations means finding the value of the unknown that makes the statement true. The golden rule is to keep the equation balanced by performing the same operation on both sides. For example, to solve 5x – 3 = 2x + 9, first collect x terms on one side: subtract 2x from both sides to get 3x – 3 = 9. Then add 3 to both sides: 3x = 12. Finally divide both sides by 3: x = 4.
解方程就是求出使等式成立的未知数的值。黄金法则是对等式两边进行相同的操作，保持方程平衡。例如，解方程 5x – 3 = 2x + 9，先把含 x 的项移到一边：两边减去 2x 得 3x – 3 = 9。然后两边加 3：3x = 12。最后两边除以 3：x = 4。
Equations involving fractions can be simplified by multiplying every term by the common denominator. Solve x/3 + 2 = 5 by multiplying through by 3: x + 6 = 15, so x = 9. Equations with brackets should be expanded first: 2(3x – 4) = 10 becomes 6x – 8 = 10, then 6x = 18, x = 3.
含有分数的方程可以通过每一项乘以公分母来简化。解 x/3 + 2 = 5，两边乘以 3 得 x + 6 = 15，所以 x = 9。有括号的方程应先展开：2(3x – 4) = 10 变成 6x – 8 = 10，然后 6x = 18，x = 3。
Sometimes you will need to form an equation yourself from a word problem. For instance, ‘I think of a number, multiply it by 4 and subtract 7, the result is 25.’ Let the number be n, so 4n – 7 = 25, giving n = 8.
有时你需要根据文字题自己建立方程。例如，“我想一个数，把它乘以 4 再减去 7，结果是 25。”设这个数为 n，那么 4n – 7 = 25，解得 n = 8。
6. Sequences and the nth Term | 序列与第 n 项
A sequence is a list of numbers following a rule. In an arithmetic sequence, the difference between consecutive terms is constant. This difference is called the common difference. The sequence 5, 9, 13, 17, … has a common difference of +4.
序列是按照某种规则排列的一列数。在等差数列中，相邻两项的差是常数，称为公差。序列 5, 9, 13, 17, … 的公差是 +4。
The nth term rule allows you to find any term in the sequence without listing them all. For an arithmetic sequence, the nth term takes the form an + b, where a is the common difference. To find b, compare the sequence to the multiples of a. For the sequence 5, 9, 13, 17, …, a = 4. The 1st term is 5, and 4 × 1 + b = 5 gives b = 1. So the nth term is 4n + 1. The 10th term is 4 × 10 + 1 = 41.
第 n 项公式让你无需列出所有项就能找到序列中的任意一项。对于等差数列，第 n 项的形式为 an + b，其中 a 是公差。要找到 b，将序列与 a 的倍数进行比较。对于序列 5, 9, 13, 17, …，a = 4。第 1 项是 5，而 4 × 1 + b = 5 得 b = 1。所以第 n 项为 4n + 1。第 10 项是 4 × 10 + 1 = 41。
Non-arithmetic sequences also appear, such as square numbers 1, 4, 9, 16, … with nth term n², or triangular numbers 1, 3, 6, 10, … with nth term n(n+1)/2. Recognising these special sequences helps in problem solving.
也会出现非等差数列，比如平方数 1, 4, 9, 16, … 的第 n 项为 n²，或者三角形数 1, 3, 6, 10, … 的第 n 项为 n(n+1)/2。识别这些特殊序列有助于解题。
7. Angles and Polygons | 角与多边形
Angle facts are built upon the straight line (sum of angles is 180°) and around a point (360°). Vertically opposite angles are equal. When two parallel lines are cut by a transversal, alternate angles are equal, corresponding angles are equal, and co-interior angles sum to 180°.
角的基础知识建立在平角（角度和为 180°）和周角（360°）之上。对顶角相等。当两条平行线被一条截线所截时，内错角相等，同位角相等，同旁内角之和为 180°。
In a triangle, the interior angles always add up to 180°. An exterior angle of a triangle equals the sum of the two opposite interior angles. Equilateral triangles have three 60° angles; isosceles triangles have two equal base angles.
三角形的内角和总是 180°。三角形的一个外角等于不相邻的两个内角之和。等边三角形的三个角都是 60°；等腰三角形的两个底角相等。
For any polygon, the sum of interior angles can be found using (n – 2) × 180°, where n is the number of sides. A pentagon has (5 – 2) × 180° = 540°. If the polygon is regular, each interior angle is that sum divided by n, e.g. a regular pentagon has 540° ÷ 5 = 108° per angle.
对于任意多边形，内角和可以用公式 (n – 2) × 180° 计算，其中 n 是边数。五边形的内角和为 (5 – 2) × 180° = 540°。如果多边形是正多边形，每个内角等于内角和除以 n，例如正五边形每个内角为 540° ÷ 5 = 108°。
8. Perimeter, Area and Volume | 周长、面积与体积
Perimeter is the distance around the outside of a shape. For a rectangle, P = 2(l + w) or simply add all side lengths. Composite shapes require careful addition of outer edges.
周长是图形外边界的长度总和。对于矩形，P = 2(l + w) 或直接加上所有边长。复合图形需要仔细地将外边界相加。
Area of a rectangle is length × width. A triangle is half of a rectangle: A = ½ × base × height. The area of a parallelogram is base × perpendicular height, and a trapezium is ½(a + b)h, where a and b are the parallel sides and h is the perpendicular height.
矩形的面积是长 × 宽。三角形面积是矩形的一半：A = ½ × 底 × 高。平行四边形面积是底 × 垂直高，梯形面积是 ½(a + b)h，其中 a 和 b 是平行边，h 是垂直高。
For circles, circumference C = πd or 2πr, and area A = πr². At this stage, π is often taken as 3.14 or answers are left in terms of π. Composite areas can be found by splitting the shape into simpler parts.
对于圆，周长 C = πd 或 2πr，面积 A = πr²。现阶段 π 通常取 3.14 或答案保留 π。复合面积可以通过将图形分解为简单部分来求得。
Volume of a cuboid is length × width × height. The volume of a prism is the area of its cross-section × length. For example, a triangular prism with cross-sectional area 10 cm² and length 8 cm has volume 10 × 8 = 80 cm³.
长方体的体积是长 × 宽 × 高。棱柱的体积是横截面积 × 长度。例如，横截面积为 10 cm²、长为 8 cm 的三棱柱的体积为 10 × 8 = 80 cm³。
9. Statistics and Averages | 统计与平均数
Data can be displayed in bar charts, pie charts, and line graphs. In Year 8, you learn to interpret grouped frequency tables and draw conclusions. The modal class is the interval with the highest frequency.
数据可以用条形图、饼图和折线图表示。在 8 年级，你学习解读分组频数表并得出结论。众数级别是频数最高的区间。
The mean is calculated by adding all values and dividing by the number of values. The median is the middle value when data are ordered; if there are two middle numbers, take their average. The mode is the most frequent value, and the range is the difference between the largest and smallest values. The range measures spread, while the mean, median and mode measure central tendency.
平均数的计算方法是将所有数值相加然后除以数值的个数。中位数是数据排序后位于中间的值；如果有两个中间数，则取它们的平均值。众数是出现频率最高的值，极差是最大值与最小值的差。极差衡量离散程度，而平均数、中位数和众数衡量数据的集中趋势。
For grouped data, the mean is estimated by taking the midpoint of each interval, multiplying by the frequency, summing these products, and dividing by the total frequency.
对于分组数据，估算平均数需要取每个区间的中点，乘以频数，将这些乘积相加，然后除以总频数。
10. Probability | 概率
Probability measures how likely an event is to happen, on a scale from 0 (impossible) to 1 (certain). The probability of an event = number of favorable outcomes / total number of possible outcomes. For a fair six-sided dice, the probability of rolling an even number is 3/6 = 1/2.
概率衡量事件发生的可能性，范围从 0（不可能）到 1（一定发生）。事件的概率 = 有利结果的数量 / 所有可能结果的总数。对于一枚均匀的六面骰子，掷出偶数的概率为 3/6 = 1/2。
The sum of probabilities of all mutually exclusive outcomes is 1. If the probability of winning a game is 0.3, the probability of not winning is 1 – 0.3 = 0.7. Two events are mutually exclusive if they cannot happen at the same time, like rolling a 3 and a 5 on a single roll.
所有互斥结果的概率之和为 1。如果赢得一场游戏的概率是 0.3，那么没赢的概率就是 1 – 0.3 = 0.7。如果两个事件不可能同时发生，则它们互斥，例如一次掷骰子同时掷出 3 和 5。
Sample space diagrams and two-way tables help list all possible outcomes for two events, making it easier to calculate probabilities. If you flip two coins, the sample space {HH, HT, TH, TT} shows that P(two heads) = 1/4.
样本空间图和双向表有助于列出两个事件所有可能的结果，使概率计算更容易。如果抛两枚硬币，样本空间 {HH, HT, TH, TT} 显示出现两个正面的概率 P = 1/4。
11. Coordinates and Transformations | 坐标与图形变换
Points are plotted using (x, y) coordinates in all four quadrants. The x-coordinate gives the horizontal movement from the origin; the y-coordinate gives the vertical movement. In the second quadrant, x is negative and y is positive.
点在四个象限中用坐标 (x, y) 标示。x 坐标表示从原点出发的水平移动；y 坐标表示垂直移动。在第二象限，x 为负，y 为正。
Transformations include translation, reflection, rotation, and enlargement. A translation moves a shape by a vector: for example, shape A translated by column vector (3, –2) moves 3 units right and 2 units down. Reflection requires a mirror line, such as y = x or the x-axis. Rotation is described by angle, direction (clockwise or anticlockwise), and centre of rotation.
图形变换包括平移、反射、旋转和缩放。平移通过一个列向量移动图形：例如图形 A 按列向量 (3, –2) 平移意味着向右移动 3 个单位，向下移动 2 个单位。反射需要一个镜像线，如 y = x 或 x 轴。旋转由角度、方向（顺时针或逆时针）和旋转中心来描述。
Enlargement changes the size of a shape by a scale factor. If the centre of enlargement is the origin and the scale factor is 2, every coordinate is multiplied by 2. Distances from the centre remain in proportion. Fractional scale factors (e.g. ½) make shapes smaller.
缩放通过比例因子改变图形的大小。如果缩放中心是原点，比例因子为 2，那么每个坐标都乘以 2。各点到中心的距离保持比例。分数比例因子（如 ½）使图形缩小。
12. Real-life and Multi-step Problems | 实际应用与多步骤问题
Word problems in the 8H book often require you to combine several mathematical skills. For example, a question might ask for the cost of carpeting a room given its dimensions and the price per square metre. You would calculate the area, then multiply by the cost per unit area, possibly adding a percentage for fitting.
8H 作业本中的文字题通常需要你综合运用多种数学技能。例如，一个问题可能会要求根据房间的尺寸和每平方米的价格计算铺地毯的费用。你需要先计算面积，然后乘以单位面积价格，可能还要加上安装费用的百分比。
Another typical problem involves interpreting timetables or converting between units (metres to centimetres, hours to minutes) while applying rates. If a tap fills a tank at 5 litres per minute, how long to fill a 1.2 m³ tank? First convert 1.2 m³ to litres (1 m³ = 1000 litres, so 1200 litres), then divide by the rate: 1200 ÷ 5 = 240 minutes = 4 hours.
另一类典型问题涉及解读时间表或单位换算（米换算为厘米，小时换算为分钟）同时应用速率。如果一个水龙头每分钟注入 5 升水，灌满一个 1.2 m³ 的水池需要多长时间？首先将 1.2 m³ 转换为升（1 m³ = 1000 升，所以是 1200 升），然后除以速率：1200 ÷ 5 = 240 分钟 = 4 小时。
Developing a systematic approach—reading the problem carefully, identifying known and unknown quantities, choosing operations, and checking the answer—is just as important as calculation fluency. Practice with multi-step problems will strengthen both your reasoning and confidence.
培养系统的方法——仔细读题、识别已知量和未知量、选择运算并检查答案——与计算的流利度同样重要。通过多步骤问题的练习，你的推理能力和自信心都会得到增强。
Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com
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