📚 Year 9 Essential Maths 9H Homework Companion | 九年级 Essential Maths 9H 作业书精要
This revision guide summarises the key topics found in the Year 9 Essential Maths 9H Homework Book. It is designed to help you consolidate higher-tier skills through clear explanations, worked examples and useful formulas. Each section pairs an English explanation with a Chinese translation, so you can study in either language or check your understanding of the terminology.
本复习指南总结了九年级 Essential Maths 9H 作业书中的核心主题。它通过清晰的讲解、例题和实用公式帮助你巩固高阶技能。每个小节都提供中英文对照，方便你使用任一语言学习，或检查自己对术语的理解。
Everything here maps directly onto the homework exercises you are already completing, so you can read one section, then return to the matching questions and finish them with confidence. The order of the sections follows the order of the topics in the homework book, from number work through to statistics.
本文所有内容都与你在做的作业练习一一对应，你可以先读一个小节，再回到对应的题目，带着信心完成练习。小节顺序与作业书中的主题顺序一致，从数与运算一直延伸到统计。
1. Number Skills and Index Laws | 数与指数法则
In 9H, index laws are tested with negative and fractional powers, not just positive integers. You must be fluent in multiplying and dividing powers with the same base, raising a power to a power, and converting between root and fractional index forms. This underpins standard form, algebraic simplification and sequences.
在 9H 课程中，指数法则不仅考查正整数幂，还涉及负指数和分数指数。你必须熟练进行同底数幂的乘除、幂的乘方，以及根式与分数指数之间的互化。这些技能是科学记数法、代数化简和数列部分的重要基础。
The four core rules are best memorised as one block, because exam questions rarely test only one of them at a time. Notice that the first two rules only work when the bases are identical.
四条核心法则最好整块记忆，因为考题很少只单独考查其中一条。注意前两条法则只有在底数相同时才成立。
aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ a⁰ = 1 a⁻ⁿ = 1 ÷ aⁿ
Fractional indices link powers to roots: a1/2 = √a, a1/3 = ∛a, and in general am/n = ⁿ√aᵐ. When you evaluate something like 272/3, take the root first and the power second, because the numbers stay small and the arithmetic is safer.
分数指数把幂与根号连接起来：a1/2 = √a，a1/3 = ∛a，一般地 am/n = ⁿ√aᵐ。计算如 272/3 这样的式子时，先开方再乘方，因为数字较小、运算更安全。
- Worked example: 2⁻³ × 2⁵ = 2⁽⁻³⁺⁵⁾ = 2² = 4
- Worked example: 272/3 = (∛27)² = 3² = 9
- Worked example: (3x²)³ = 27x⁶ (the power applies to the coefficient as well)
- 例题：2⁻³ × 2⁵ = 2⁽⁻³⁺⁵⁾ = 2² = 4
- 例题：272/3 = (∛27)² = 3² = 9
- 例题：(3x²)³ = 27x⁶（指数同时作用在系数上）
Alongside indices, the number section of the homework book keeps returning to prime factorisation, HCF and LCM, and arithmetic with negative numbers. Write every number as a product of primes first; HCF is then the product of the shared primes, and LCM is the product of the highest power of every prime that appears.
除了指数，作业书的“数”部分还反复考查质因数分解、最大公因数与最小公倍数，以及负数运算。先把每个数写成质数乘积；最大公因数就是公共质数的乘积，最小公倍数则是所有出现过的质数最高次幂的乘积。
2. Standard Form, Surds and Rounding | 科学记数法、根式与近似值
. Large numbers need a positive index, small numbers a negative one. To multiply in standard form, multiply the A values and add the indices; to divide, divide the A values and subtract the indices, then adjust if the new A value is no longer between 1 and 10.
科学记数法把一个数写成 A × 10ⁿ，其中 1 ≤ A < 10，n 为整数。大数用正指数，小数用负指数。相乘时把 A 值相乘、指数相加；相除时把 A 值相除、指数相减，若新的 A 值不再介于 1 与 10 之间，还要再做调整。
Surds appear whenever a square root cannot be simplified to a whole number. The key rules are √ab = √a × √b and √a ÷ b = √a ÷ √b. To simplify √48, look for the largest square factor: √48 = √16 × √3 = 4√3.
当平方根无法化成整数时就会出现根式。核心法则是 √ab = √a × √b 以及 √a ÷ b = √a ÷ √b。化简 √48 时找出最大的平方因数：√48 = √16 × √3 = 4√3。
√a × √b = √ab √a ÷ √b = √a ÷ b 1 ÷ √2 = √2 ÷ 2
Rationalising the denominator means rewriting a fraction so that no surd is left underneath. Multiply the top and bottom by the surd on the bottom, then simplify. Higher-tier questions also combine surds with expanding brackets, for example (3 + √2)(3 − √2) = 9 − 2 = 7.
分母有理化就是把分数改写成分母中不含根号的形式。把分子分母同时乘以分母上的根式，然后化简。高阶题目还会把根式与展开括号结合，例如 (3 + √2)(3 − √2) = 9 − 2 = 7。
n as 7.4 cm to one decimal place, the true value satisfies 7.35 ≤ x < 7.45. The upper bound is excluded, and writing ≤ on both sides is a common mark-losing error.
近似值题考查的是有效数字而非小数位，并且常要求写出近似值的误差区间。若某长度四舍五入到一位小数得 7.4 cm，则真实值满足 7.35 ≤ x < 7.45。上界不包含在内，两边都写 ≤ 是常见的失分错误。
3. Expanding, Factorising and Algebraic Manipulation | 展开、因式分解与代数变形
Expanding brackets and factorising are inverse processes, and the homework book moves between them constantly. Expanding (x + 3)(x − 5) gives x² − 5x + 3x − 15, which simplifies to x² − 2x − 15. Every term in the first bracket must multiply every term in the second.
展开括号与因式分解互为逆运算，作业书在这两者之间不断切换。展开 (x + 3)(x − 5) 得到 x² − 5x + 3x − 15，化简为 x² − 2x − 15。第一个括号中的每一项都必须与第二个括号中的每一项相乘。
Three special patterns are examined every year and should be recognised on sight. Square brackets give a perfect square, and brackets that differ only in sign give the difference of two squares.
每年必考的三个特殊模式应当一眼认出。完全相同的括号得到完全平方式，而仅符号不同的括号得到平方差。
+ b² (a − b)² = a² − 2ab + b² a² − b² = (a + b)(a − b)
When factorising a quadratic of the form x² + bx + c, find two numbers that multiply to c and add to b. For x² + 7x + 12 those numbers are 3 and 4, so the answer is (x + 3)(x + 4). If the coefficient of x² is not 1, the homework book expects either the grid method or trial with the factors of a and c.
分解 x² + bx + c 形式的二次式时，找出两个数，使它们的积为 c、和为 b。对 x² + 7x + 12 而言，这两个数是 3 和 4，因此答案为 (x + 3)(x + 4)。若 x² 的系数不是 1，作业书要求使用表格法，或用 a 与 c 的因数进行试算。
Always check for a common factor before anything else: 3x² + 6x factorises fully as 3x(x + 2), and leaving it as x(3x + 6) would lose a mark. Algebraic fractions follow the same rule, so you must factorise top and bottom before you cancel.
动手之前永远先看有没有公因式：3x² + 6x 的完全分解是 3x(x + 2)，若写成 x(3x + 6) 会丢分。代数分式遵循同样的规则，必须先对分子分母分解因式，再约分。
4. Linear Equations, Inequalities and Formulae | 一元方程、不等式与公式变形
Solving a linear equation means doing the same operation to both sides until the unknown stands alone. Clear fractions first by multiplying every term by the common denominator, expand any brackets, then collect the unknowns on one side and the numbers on the other.
解一元方程就是对两边做相同的运算，直到未知数单独留在一边。先乘以公分母消去分数，再展开括号，然后把未知数集中到一边、常数集中到另一边。
sign. Always represent the final answer on a number line, using an open circle for < and > and a filled circle for ≤ and ≥.
不等式的解法完全相同，只有一个关键例外：乘以或除以负数时，不等号方向要反转。最后的答案一定要画在数轴上，< 与 > 用空心圆，≤ 与 ≥ 用实心圆。
. Multiply through by 6 to obtain 2(2x + 1) = 3(x − 4), expand, and solve to find x = −14.
典型的高阶题目会把分数与两边都有未知数结合起来，例如 (2x + 1) ÷ 3 = (x − 4) ÷ 2。两边同乘 6 得 2(2x + 1) = 3(x − 4)，展开后解得 x = −14。
(y + 7) ÷ 3. If the unknown appears twice, collect both terms first and factorise it out.
公式变形就是用字母代替数字的同一套技能。要使 x 成为 y = 3x − 7 的主元，两边加 7 再除以 3，得 x = (y + 7) ÷ 3。若未知数出现两次，先把它集中，再提取公因式。
Simultaneous equations also belong to this section. Use elimination when the coefficients are close, and substitution when one equation already gives one letter in terms of the other. Always substitute your answer back to check both equations are satisfied.
联立方程也属于这一小节。当系数较接近时使用消元法，当其中一个方程已给出某字母关于另一字母的表达式时使用代入法。求出答案后一定要代回两个方程检验是否都成立。
5. Sequences and nth Term | 数列与通项公式
Every sequence in the homework book is either linear, quadratic or one of the special families such as square, cube, triangular and Fibonacci numbers. Begin by finding the difference between consecutive terms, because that tells you which type you are dealing with.
作业书中的数列要么是线性数列，要么是二次数列，或者属于平方数、立方数、三角数、斐波那契数列等特殊类型。先求相邻两项的差，因为差值能告诉你它属于哪一类。
he difference is 3, so the first part is 3n; testing n = 1 gives 3, which is 2 less than 5, so the nth term is 3n + 2.
对一阶差分为常数的线性数列，通项公式形如 an + b，其中 a 是公差。对数列 5, 8, 11, 14，公差为 3，所以先写出 3n；代入 n = 1 得 3，比 5 少 2，所以通项为 3n + 2。
For a quadratic sequence the first differences are not constant but the second differences are. Halve the second difference to get the coefficient of n², subtract n² terms from the original sequence, and the remainder is a linear sequence whose nth term you already know how to find.
二次数列的一阶差分不是常数，但二阶差分是常数。把二阶差分除以 2，得到 n² 的系数；再从原数列中减去 n² 项，剩下的部分就是一个你已经会求通项的线性数列。
nth term of a quadratic sequence = (½ × second difference)n² + linear remainder
- Special sequences to recognise: squares 1, 4, 9, 16, 25; cubes 1, 8, 27, 64; triangular 1, 3, 6, 10, 15
- 需记住的特殊数列：平方数 1、4、9、16、25；立方数 1、8、27、64；三角数 1、3、6、10、15
Sequences also connect to practical problems, such as patterns of matchsticks or tiles where you must describe the number of items in the nth diagram. Draw the first three diagrams, tabulate the values, and treat it as an nth term question.
数列还会与实际情境结合，例如火柴棒或方砖的排列，需要你描述第 n 个图形中物品的数量。先画出前三个图形，把数值列成表，再当作通项问题处理。
6. Ratio, Proportion and Percentages | 比、比例与百分数
Ratio problems are solved by finding the value of one part. To divide £96 in the ratio 3 : 5, add the parts to get 8, divide 96 by 8 to find one part is 12, then multiply: 36 and 60. Stating what one part represents is the step most students skip and most marks depend on.
比例问题的解法是先求出一份的值。把 96 英镑按 3 : 5 分配，先把份数相加得 8，用 96 ÷ 8 得一份为 12，再相乘得到 36 和 60。说明“一份代表多少”是多数学生省略、却是得分关键的一步。
Direct proportion means both quantities grow together, so the ratio stays fixed. Inverse proportion means that as one quantity increases the other decreases, so the product stays constant. In both cases, find the value for one unit first, then scale to the required amount.
正比例指两个量同向变化，比值保持不变。反比例指一个量增大时另一个量减小，乘积保持不变。两种情况下都先求“单位量”的值，再按需求放大。
ercentages and compound interest. For a percentage change, use the multiplier method: a 15% increase is × 1.15 and a 15% decrease is × 0.85.
9H 的百分数内容包括百分数的增减、逆百分数与复利。计算百分比变化时使用乘数法：增加 15% 是 × 1.15，减少 15% 是 × 0.85。
Final amount = Original × (1 ± r ÷ 100)ⁿ
ions never start by subtracting the percentage from the final value. If a price after a 20% discount is £64, then £64 represents 80% of the original, so the original price is 64 ÷ 0.8 = £80. Compound interest uses the same multiplier raised to the power of the number of years.
逆百分数问题绝不能直接从最终值中减去百分比。若某商品打八折后为 64 英镑，则 64 英镑是原价的 80%，原价为 64 ÷ 0.8 = 80 英镑。复利使用同一乘数，指数为年数。
7. Straight-Line Graphs and Coordinates | 直线图像与坐标
t and c is the y-intercept. To plot a line, build a table of values, plot at least three points, and check that they lie in a straight line before drawing with a ruler.
直线方程写作 y = mx + c，其中 m 是斜率，c 是纵截距。画直线时先列出数值表，描出至少三个点，确认它们共线后再用直尺连线。
Gradient measures steepness and can be calculated from any two points on the line. A positive gradient rises from left to right; a negative gradient falls.
斜率表示倾斜程度，可用直线上任意两点求出。正斜率从左向右上升，负斜率从左向右下降。
m = (y₂ − y₁) ÷ (x₂ − x₁)
To find the equation of a line through
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