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Year 9 OCR Further Maths: Formula & Theorem Quick Reference | 九年级OCR进阶数学:公式定理速查手册

2026-07-18 11:35 · 本文部分内容由AI辅助创作

📚 Year 9 OCR Further Maths: Formula & Theorem Quick Reference | 九年级OCR进阶数学:公式定理速查手册

This quick reference handbook compiles essential formulas, identities and theorems for the Year 9 OCR Further Maths course. Each entry is presented in a concise, exam-focused style to support rapid revision and classroom learning.

本速查手册汇编了九年级OCR进阶数学课程的核心公式、恒等式与定理,所有条目均以简明、贴近考试的风格呈现,便于快速复习与课堂巩固。


1. Algebraic Expressions and Identities | 代数表达式与恒等式

efficients. The distributive law a(b + c) = ab + ac is fundamental for expanding brackets.

化简代数表达式时,同类项可以通过系数相加减合并。分配律 a(b + c) = ab + ac 是展开括号的基础法则。

a(b + c) = ab + ac

x² + (a+b)x + ab. Special products include perfect squares and difference of two squares.

两个二项式相乘时使用FOIL法则展开:(x + a)(x + b) = x² + (a+b)x + ab 。常见特殊展开包括完全平方公式和平方差公式。

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

a² − b² = (a + b)(a − b)

Factorising an expression is the reverse of expanding. Always look for a common factor first, then recognise any standard forms such as difference of squares or quadratic trinomials.

因式分解是展开的逆运算。应先提取公因子,再识别标准形式,如平方差或二次三项式。


2. Linear and Simultaneous Equations | 线性方程与联立方程

A linear equation can be solved by isolating the unknown. Use inverse operations: addition/subtraction then multiplication/division, keeping the equation balanced.

解一元一次方程可通过逆运算求解未知数:先加减后乘除,保持方程两边平衡。

m is the gradient and c is the y-intercept. The general form ax + by + c = 0 is also used.

直线方程通常写作 y = mx + c ,其中 m 为斜率,c 为y轴截距。也使用一般式 ax + by + c = 0 。

m = (y₂ − y₁) / (x₂ − x₁)

Simultaneous linear equations in two variables can be solved by elimination or substitution. Multiplying equations to obtain equal coefficients often helps eliminate one variable.

含两个变量的联立一次方程可通过消元法或代入法求解。通常先将方程乘上适当倍数,使某个变量系数相等,再消元。

The solution to a pair of simultaneous equations represents the coordinates where their graphs intersect. Parallel lines have no solution, and coincident lines have infinitely many.

联立方程组的解代表两条直线图像的交点坐标。平行线无解,重合直线有无穷多解。


3. Quadratic Equations | 二次方程

0. It can be solved by factorising, completing the square, or using the quadratic formula.

二次方程的形式为 ax² + bx + c = 0 (a ≠ 0)。可通过因式分解、配方法或求根公式求解。

x = (−b ± √b² − 4ac) / (2a)

two real distinct roots if Δ > 0, one repeated root if Δ = 0, and no real roots if Δ < 0.

根号内的表达式 b² − 4ac 称为判别式 (Δ)。它决定根的性质:Δ > 0 时有两个相异实根,Δ = 0 时有一个重根,Δ < 0 时无实根。

rewrites the quadratic as a(x + p)² + q = 0, revealing the vertex of its graph at (−p, q).

配方法将二次式写成 a(x + p)² + q = 0 的形式,其图像顶点为 (−p, q)。

ic can be expressed as (px + q)(rx + s) = 0, then the solutions are x = −q/p and x = −s/r.

优先尝试因式分解:若二次式可化为 (px + q)(rx + s) = 0 ,则解为 x = −q/p 和 x = −s/r 。


4. Inequalities | 不等式

. Solving them follows the same steps as linear equations, but multiplying or dividing by a negative number reverses the inequality sign.

不等式使用 >、<、≥ 和 ≤ 符号。其求解步骤与一次方程相同,但当乘以或除以负数时,不等号方向要反转。

osed circle for inclusive ones. Compound inequalities like a < x ≤ b describe an interval.

解集可用数轴表示:严格不等式用空心圆圈,包含等号的用实心圆点。复合不等式如 a < x ≤ b 描述一个区间。

To solve quadratic inequalities, first solve the corresponding equation, then test intervals between the critical values to determine where the inequality holds.

解二次不等式时,先求解对应的方程,然后在临界值划分的各区间内检验符号,确定不等式成立的范围。


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5. Sequences and Series | 序列与级数

ommon difference d. Its nth term is given by uₙ = a + (n − 1)d, where a is the first term.

等差数列以固定的公差 d 递增或递减。第 n 项的通项公式为 uₙ = a + (n − 1)d ,其中 a 为首项。

e is Sₙ = n/2 [2a + (n − 1)d] or, equivalently, Sₙ = n/2 (a + l) where l is the last term.

等差数列前 n 项的和为 Sₙ = n/2 [2a + (n − 1)d] ,或等价地 Sₙ = n/2 (a + l) ,其中 l 为末项。

Sₙ = n/2 (a + l)

Recognising whether a sequence is arithmetic often requires checking that the difference between consecutive terms remains constant. Patterns from diagrams can also generate sequences.

判断数列是否为等差数列通常需检查连续项之差是否恒定。根据图形规律也可生成数列。


6. Indices and Standard Form | 指数与标准型

The laws of indices simplify calculations with powers. They apply when bases are the same and the operation is multiplication, division or raising a power to another power.

指数法则用于简化幂运算,要求底数相同,可用于乘法、除法或幂的乘方。

aᵐ × aⁿ = aᵐ⁺ⁿ同底数幂相乘,指数相加
aᵐ ÷ aⁿ = aᵐ⁻ⁿ同底数幂相除,指数相减
(aᵐ)ⁿ = aᵐⁿ幂的乘方,指数相乘
a⁰ = 1 (a ≠ 0)任何非零数的0次幂等于1
a⁻ⁿ = 1 / aⁿ负指数表示倒数

where 1 ≤ a < 10 and n is an integer. It is used for very large or very small quantities.

标准型将一个数记为 a × 10ⁿ ,其中 1 ≤ a < 10 且 n 为整数。多用于很大或很小的数。

Calculations in standard form combine the decimal parts and the powers of ten separately, then adjust to keep the coefficient between 1 and 10.

在标准型下进行运算时,先将小数部分与10的幂次分别处理,再调整系数使其介于1与10之间。


7. Coordinate Geometry | 坐标几何

The distance between two points (x₁, y₁) and (x₂, y₂) is found using the distance formula, which is derived from Pythagoras' theorem.

两点 (x₁, y₁) 和 (x₂, y₂) 间的距离由距离公式求出,该公式由勾股定理推导而来。

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

The midpoint M of a segment joining the two points has coordinates given by the average of the x-coordinates and the y-coordinates.

连接两点的线段中点 M 的坐标为 x 坐标与 y 坐标的平均值。

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

ent m of the line through (x₁, y₁) and (x₂, y₂) is rise over run: m = (y₂ − y₁)/(x₂ − x₁).

过 (x₁, y₁) 和 (x₂, y₂) 的直线斜率 m 为纵坐标之差除以横坐标之差:m = (y₂ − y₁)/(x₂ − x₁) 。

adients. Two lines with gradients m₁ and m₂ are perpendicular if and only if m₁ × m₂ = −1.

平行线斜率相等。斜率分别为 m₁ 和 m₂ 的两条直线垂直当且仅当 m₁ × m₂ = −1 。


8. Pythagoras and Trigonometry | 勾股定理与三角学

In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides.

在直角三角形中,斜边(直角对边)的平方等于两直角边的平方和。

a² + b² = c²

f three sides satisfy a² + b² = c², the triangle is right-angled with c as the hypotenuse.

逆定理同样成立:若三边满足 a² + b² = c² ,则该三角形为直角三角形,c 为斜边。

Trigonometric ratios in a right-angled triangle relate an acute angle to the ratios of two sides. Use SOH CAH TOA to recall them:

直角三角形中的三角比将锐角与两边之比建立联系。使用 SOH CAH TOA 帮助记忆:

sin θ = opposite / hypotenuse

cos θ = adjacent / hypotenuse

tan θ = opposite / adjacent

Exact values for special angles should be known for quick problem-solving:

需熟记特殊角的精确值以便快速解题:

sin 30° = 1/2sin 45° = √2/2sin 60° = √

Published by TutorHao | Year 9 进阶数学 Revision Series | aleveler.com

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