Edexcel A-Level Pure Maths: Core Techniques and Worked Examples | 爱德思 A-Level 纯数学：核心方法与例题

📚 Edexcel A-Level Pure Maths: Core Techniques and Worked Examples | 爱德思 A-Level 纯数学：核心方法与例题

This revision article brings together the core pure-mathematics techniques that regularly appear in Edexcel A-Level Mathematics Paper 1 and Paper 2. It is designed around the mixed practice page 'pdfjoiner (4)-076' and covers algebra, functions, trigonometry, calculus, sequences, coordinate geometry and vectors. Every section pairs the technique with a worked example so that you can see exactly how the marks are earned, and the Pure 1 and Pure 2 papers each carry 100 marks and 33⅓% of the qualification, so fluency here is not optional.

本篇复习文章汇集了爱德思 A-Level 数学 Paper 1 和 Paper 2 中经常出现的纯数学核心方法。文章围绕综合练习页 'pdfjoiner (4)-076' 设计，涵盖代数、函数、三角、微积分、数列、坐标几何与向量等内容。每一节都把方法配上例题，让你清楚看到分数是如何拿到的；Pure 1 与 Pure 2 各占 100 分、占总成绩 33⅓%，所以这些基本功必须熟练。

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1. Algebraic Simplification, Surds and Indices | 代数化简、根式与指数

When simplifying surds, always look for square factors and rationalise denominators where possible. The two rules you must know are that the square root of a product equals the product of the square roots, and the same for a quotient.

化简根式时，应始终寻找平方因子，并尽可能有理化分母。必须掌握的两条法则是：积的平方根等于平方根的积，商也同样成立。

√a × √b = √ab　　√a ÷ b = √a ÷ √b

For example, √48 can be written as √16 × 3 = √16 × √3 = 4√3. Always pull out the largest square factor you can find, not just any factor, otherwise you will have to simplify twice.

例如，√48 可以写成 √16 × 3 = √16 × √3 = 4√3。要一次提取最大的平方因子，而不是随便提一个，否则你需要化简两次。

To rationalise a denominator such as 1/(√3 + √2), multiply the numerator and denominator by the conjugate √3 - √2. The surds cancel in the denominator because of the difference of two squares.

要化简 1/(√3 + √2) 这样的分母，将分子和分母同时乘以共轭根式 √3 - √2。由于平方差公式，分母中的根式会消掉。

1/(√3 + √2) × (√3 - √2)/(√3 - √2) = (√3 - √2)/(3 - 2) = √3 - √2

The same conjugate trick works with a single surd: 1/(3 + √5) × (3 - √5)/(3 - √5) = (3 - √5)/(9 - 5) = (3 - √5)/4.

同样的共轭技巧也适用于单项根式：1/(3 + √5) × (3 - √5)/(3 - √5) = (3 - √5)/(9 - 5) = (3 - √5)/4。

Index laws are examined constantly, often disguised inside calculus questions. Learn them as a single family rather than as separate facts.

指数法则被反复考查，而且常常伪装在微积分题目里。请把它们当作一个整体来记忆，而不是零散的知识点。

• aᵐ × aⁿ = aᵐ⁺ⁿ　and　aᵐ ÷ aⁿ = aᵐ⁻ⁿ
• (aᵐ)ⁿ = aᵐⁿ　and　(ab)ⁿ = aⁿbⁿ
• a⁰ = 1　and　a⁻ⁿ = 1 ÷ aⁿ
• a^m/n = (ⁿ√a)ᵐ = ⁿ√aᵐ

Worked example: 16^3/4 = (⁴√16)³ = 2³ = 8, and (8x⁶)^2/3 = 8^2/3 × x⁴ = 4x⁴. A very common error is writing √a + b = √a + √b, which is false: √9 + 16 = 5, but √9 + √16 = 7.

例题：16^3/4 = (⁴√16)³ = 2³ = 8，而 (8x⁶)^2/3 = 8^2/3 × x⁴ = 4x⁴。一个非常常见的错误是写成 √a + b = √a + √b，这是错的：√9 + 16 = 5，但 √9 + √16 = 7。

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2. Algebraic Fractions and Partial Fractions | 代数分式与部分分式

In algebraic fractions, factorise both numerator and denominator first, then cancel common factors. Nothing may be cancelled until it is a factor of the whole expression, never a single term.

在代数分式中，应先对分子和分母因式分解，然后约去公因式。只有当一个整体是因子时才能约去，绝不能只约去某一项。

9) ÷ (x² + x - 12) = (x - 3)(x + 3) ÷ ((x + 4)(x - 3)) = (x + 3) ÷ (x + 4), x ≠ 3, x ≠ -4

If the fraction is improper, meaning the degree of the top is greater than or equal to the degree of the bottom, divide first and then deal with the remainder.

如果分式是假分式，即分子的次数大于或等于分母的次数，应先做除法，再处理余项。

Partial fractions reverse this process. For a proper fraction with distinct linear factors, write the expression as a sum of simple fractions with unknown constants A and B.

部分分式则把这个过程反过来。对于分母为互异一次因式的真分式，把它写成带未知常数 A 与 B 的简单分式之和。

(4x + 5) ÷ ((x - 1)(x + 2)) = A ÷ (x - 1) + B ÷ (x + 2)

bstituting x = -2 gives -3 = -3B, so B = 1. Therefore the answer is 3/(x - 1) + 1/(x + 2).

两边同乘分母得到 4x + 5 = A(x + 2) + B(x - 1)。代入 x = 1 得 9 = 3A，所以 A = 3；代入 x = -2 得 -3 = -3B，所以 B = 1。因此答案为 3/(x - 1) + 1/(x + 2)。

For a repeated factor, one bracket is needed for each power. For example, (x + 5)/(x(x + 1)²) decomposes into A/x + B/(x + 1) + C/(x + 1)². Partial fractions are most often used later for integrating rational functions.

若分母含重因式，则每一个幂次都要写一个括号项。例如 (x + 5)/(x(x + 1)²) 分解为 A/x + B/(x + 1) + C/(x + 1)²。部分分式最常见的后续用途是对有理函数积分。

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3. Quadratic Functions, Discriminant and Completing the Square | 二次函数、判别式与配方法

y Δ = b² - 4ac. It tells us how many real roots the equation has without solving it fully.

二次方程 ax² + bx + c = 0 的判别式为 Δ = b² - 4ac。它使我们无需完整求解方程即可判断实根的个数。

Value of Δ	Nature of roots	Graph
Δ > 0	two distinct real roots	crosses the x-axis twice
Δ = 0	one repeated real root	touches the x-axis
Δ < 0	no real roots	does not meet the x-axis

f k for which x² + kx + 9 = 0 has equal roots, set Δ = 0: k² - 36 = 0, so k = 6 or k = -6.

例如，x² - 5x + 3 = 0 的判别式 Δ = (-5)² - 4(1)(3) = 25 - 12 = 13，因此它有两个不同实根。若题目问 x² + kx + 9 = 0 有等根时 k 的取值，令 Δ = 0：k² - 36 = 0，所以 k = 6 或 k = -6。

Completing the square transforms a quadratic into the form a(x + p)² + q, which reveals the vertex (-p, q) and the line of symmetry

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