📚 1 Introduction to Algebraic Expressions | 代数表达式导论
Algebraic expressions form the language of A-Level Mathematics. In this opening chapter, you will consolidate and extend the algebraic skills encountered at GCSE, building a toolkit that underpins everything from calculus to mechanics. The focus is on manipulating expressions confidently: applying index laws, expanding brackets, factorising polynomials, and working with surds and algebraic fractions. Mastering these fundamentals early will save you time and reduce errors in more advanced topics.
代数表达式是 A-Level 数学的语言。在这一开篇章节中，你将巩固并拓展在 GCSE 阶段接触过的代数技能，构建起支撑从微积分到力学等所有内容的工具包。重点在于自信地操作表达式：运用指数法则、展开括号、因式分解多项式、处理根式（surd）以及代数分式。尽早掌握这些基础知识将为你在更高级的专题中节省时间并减少错误。
1. The Index Laws | 指数运算法则
Indices (also called powers or exponents) are a shorthand for repeated multiplication. The three fundamental index laws for any real base a and integers m, n are: am × an = am+n, am ÷ an = am−n, and (am)n = am×n. These rules only apply when the base is the same. At A-Level, you are expected to apply them fluently to both numeric and algebraic bases, and to combine them with other operations without hesitation.
指数（也称幂或乘方）是重复相乘的简写形式。对于任意实数底 a 和整数 m、n，三条基本指数法则是：am × an = am+n，am ÷ an = am−n，以及 (am)n = am×n。这些法则仅在底数相同时适用。在 A-Level 阶段，你需要熟练地将它们应用于数值底和代数底，并能毫不迟疑地与其他运算结合使用。
2. Negative and Fractional Indices | 负指数与分数指数
A negative index indicates a reciprocal: a−n = 1 / an (a ≠ 0). This definition is consistent with the index laws; for example, am ÷ am+n = a−n yields the same result. Fractional indices represent roots: a1/n = ⁿ√a (the nth root of a), and am/n = (ⁿ√a)m = ⁿ√(am). The order of taking the root and the power does not matter. Edexcel exam questions frequently test your ability to rewrite expressions like 1/x² as x−2 or ∛x as x1/3 before differentiating or integrating.
负指数表示倒数：a−n = 1 / an（a ≠ 0）。该定义与指数法则一致；例如 am ÷ am+n = a−n 会得到相同结果。分数指数则表示方根：a1/n = ⁿ√a（a 的 n 次方根），以及 am/n = (ⁿ√a)m = ⁿ√(am)。先开方还是先乘方并不影响结果。Edexcel 试卷经常考查你是否能在求导或积分之前，将类似 1/x² 的式子改写为 x−2，或将 ∛x 改写为 x1/3。
3. Expanding Single Brackets | 展开单个括号
Expanding means removing brackets by multiplying each term inside by the factor outside. For a single bracket: a(b + c) = ab + ac. With a negative factor, signs must be handled carefully: −2(3x − 5) = −6x + 10. When the multiplier is an algebraic term, apply index laws: x²(3x − 4) = 3x³ − 4x². At A-Level, single bracket expansions often appear inside larger expressions, so speed and accuracy are essential.
展开是指将括号内的每一项乘以外部的因子来去掉括号。对于单个括号：a(b + c) = ab + ac。若因子为负，必须小心处理符号：−2(3x − 5) = −6x + 10。当乘数是代数项时，需运用指数法则：x²(3x − 4) = 3x³ − 4x²。在 A-Level 中，单个括号的展开经常出现在更大的表达式内部，因此速度和准确性至关重要。
4. Expanding Double Brackets | 展开两个括号
To expand a product of two binomials such as (x + a)(x + b), multiply each term in the first bracket by each term in the second, then simplify: (x + a)(x + b) = x² + bx + ax + ab = x² + (a + b)x + ab. A common extension is the square of a binomial: (x + a)² = x² + 2ax + a². Edexcel questions may ask you to expand and simplify expressions like (2x − 3)(x + 4) or (x + 2)(x² − x + 1), where the second bracket has three terms. The principle remains the same: multiply every term in the first bracket by every term in the second.
要展开两个二项式的乘积，如 (x + a)(x + b)，需将第一个括号中的每一项分别乘以第二个括号中的每一项，然后化简：(x + a)(x + b) = x² + bx + ax + ab = x² + (a + b)x + ab。一个常见的拓展是二项式的平方：(x + a)² = x² + 2ax + a²。Edexcel 试题可能会要求你展开并化简诸如 (2x − 3)(x + 4) 或 (x + 2)(x² − x + 1) 的式子，其中第二个括号包含三项。原理相同：将第一个括号内的每一项与第二个括号内的每一项相乘。
5. Factorising by Common Factors | 提取公因式分解
Factorising is the reverse of expanding; you rewrite an expression as a product of its factors. The simplest case is taking out the highest common factor (HCF). For example, 6x² + 9x = 3x(2x + 3). Always check that the factor you remove is the greatest possible, including any common variables with the smallest exponent. This skill becomes vital when simplifying algebraic fractions or solving equations where one side is set to zero.
因式分解是展开的逆运算；你将一个表达式重新写成其因子的乘积。最简单的情形是提取最大公因式（HCF）。例如，6x² + 9x = 3x(2x + 3)。务必确保你提取的因式是可能的最大公因式，包括最小指数次数的公共变量。当化简代数分式，或求解方程时将一边置为零，这一技能便至关重要。
6. Factorising Quadratics | 二次三项式的因式分解
A quadratic expression of the form x² + bx + c can often be factorised into (x + p)(x + q) where p + q = b and pq = c. For example, x² + 5x + 6 = (x + 2)(x + 3) because 2 + 3 = 5 and 2 × 3 = 6. When the coefficient of x² is not 1 (e.g. 2x² + 7x + 3), you may need to split the middle term or use trial and error: look for two numbers that multiply to ac (2×3=6) and add to b (7), namely 6 and 1, then factor by grouping.
形如 x² + bx + c 的二次式通常可以因式分解为 (x + p)(x + q)，其中 p + q = b 且 pq = c。例如，x² + 5x + 6 = (x + 2)(x + 3)，因为 2 + 3 = 5 而 2 × 3 = 6。当 x² 的系数不为 1 时（例如 2x² + 7x + 3），你可能需要拆分中间项或使用试错法：寻找两个数，其乘积为 ac（2×3=6）且和为 b（7），即 6 和 1，然后分组分解。
7. Difference of Two Squares | 平方差公式
One of the most recognisable patterns in algebra is the difference of two squares: a² − b² = (a + b)(a − b). This identity is extremely useful when factorising expressions such as x² − 9, which becomes (x + 3)(x − 3), or 4x² − 25 = (2x + 5)(2x − 5). It also appears when simplifying rational expressions and solving quadratic equations. Always check whether a subtraction of two perfect squares can be applied before trying other factorising methods.
代数中最容易识别的模式之一是平方差：a² − b² = (a + b)(a − b)。这一恒等式在因式分解诸如 x² − 9（分解为 (x + 3)(x − 3)）或 4x² − 25 = (2x + 5)(2x − 5) 的表达式时极为有用。它还会出现在化简有理式以及求解二次方程的过程中。在尝试其他因式分解方法之前，务必先检查是否可以使用两个完全平方数相减的形式。
8. Simplifying Surds | 化简根式
Surds are irrational roots that cannot be expressed as exact decimals, such as √2 or ∛5. The key rules are: √a × √b = √(ab) and √a ÷ √b = √(a/b) for a, b ≥ 0. To simplify a surd like √48, identify the largest square factor: √48 = √(16 × 3) = √16 × √3 = 4√3. You should also be able to add and subtract surds when they have the same radicand, e.g. 3√2 + 5√2 = 8√2. Surd manipulation is frequently tested alongside indices and rationalisation.
根式（surd）是无法表示为精确小数的无理方根，例如 √2 或 ∛5。关键规则是：对于 a, b ≥ 0，有 √a × √b = √(ab) 以及 √a ÷ √b = √(a/b)。要化简如 √48 的根式，需找出最大的平方因数：√48 = √(16 × 3) = √16 × √3 = 4√3。你还应能够对具有相同被开方数的根式进行加减，例如 3√2 + 5√2 = 8√2。根式的运算常与指数和有理化一同考查。
9. Rationalising the Denominator | 分母有理化
It is conventional to rewrite a fraction so that the denominator is a rational number. For a simple denominator like √c, multiply numerator and denominator by √c: 1/√c = √c / c. For a binomial denominator such as a + √b, use the conjugate a − √b because (a + √b)(a − √b) = a² − b, which is rational. For example, 1/(3 + √2) can be rationalised to (3 − √2) / (9 − 2) = (3 − √2)/7. This skill removes surds from denominators and often yields a simpler algebraic form.
习惯上我们会将分数重写，使分母成为有理数。对于像 √c 这样的简单分母，将分子和分母同时乘以 √c：1/√c = √c / c。对于 a + √b 这样的二项式分母，则使用共轭式 a − √b，因为 (a + √b)(a − √b) = a² − b，结果为有理数。例如，1/(3 + √2) 可通过有理化为 (3 − √2) / (9 − 2) = (3 − √2)/7。这一技能可消除分母中的根式，并通常能给出更简洁的代数形式。
10. Algebraic Fractions | 代数分式运算
Algebraic fractions combine the techniques of factorising, expanding, and index laws. To simplify a fraction like (x² − 4)/(x² − x − 2), factorise both numerator and denominator: (x − 2)(x + 2) / (x − 2)(x + 1), then cancel the common factor (x − 2) to obtain (x + 2)/(x + 1), provided x ≠ 2. When adding or subtracting fractions, find a common denominator; when multiplying, factorise and cancel before multiplying; when dividing, multiply by the reciprocal. Fluency with these operations is essential for success in calculus and beyond.
代数分式综合运用了因式分解、展开以及指数法则等技巧。要化简像 (x² − 4)/(x² − x − 2) 这样的分式，需将分子和分母分别因式分解：(x − 2)(x + 2) / (x − 2)(x + 1)，然后约去公因式 (x − 2) 得到 (x + 2)/(x + 1)，前提是 x ≠ 2。加减分式时，先找到公分母；乘除分式时，先因式分解并约分，然后再相乘；除法时乘以倒数。熟练运用这些运算是取得微积分及后续内容学习成功的关键。
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