Algebraic Fractions | 代数分式

📚 Algebraic Fractions | 代数分式

Algebraic fractions are rational expressions of the form P(x)/Q(x), where P and Q are polynomials and Q is not identically zero. Mastering them is essential for A-level calculus, equation solving, and partial fractions.

代数分式是形如 P(x)/Q(x) 的有理式，其中 P 和 Q 是多项式，且 Q 不恒等于零。掌握代数分式对 A-level 微积分、方程求解和部分分式都至关重要。

This revision guide covers the definition and domain restrictions, simplification by factorising, the four operations, complex fractions, equations containing fractions, improper fractions, polynomial division and partial fractions, together with the errors that cost the most marks in examinations.

本篇复习指南涵盖定义与定义域限制、因式分解化简、四则运算、繁分式、含分式的方程、假分式、多项式除法与部分分式，并总结考试中最容易失分的典型错误。

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1. Understanding Algebraic Fractions | 理解代数分式

An algebraic fraction is a quotient of two polynomials, such as (x² − 3x + 2)/(x − 1). The denominator cannot be zero, so every algebraic fraction carries a hidden domain restriction.

代数分式是两个多项式的商，例如 (x² − 3x + 2)/(x − 1)。分母不能为零，因此每个代数分式都带有隐藏的定义域限制。

An algebraic fraction is called proper if the degree of the numerator is less than the degree of the denominator, and improper if the degree of the numerator is greater than or equal to the degree of the denominator. Improper fractions must normally be written as a polynomial plus a proper fraction before you can integrate them or split them into partial fractions.

若分子的次数小于分母的次数，该分式称为真分式；若分子的次数大于或等于分母的次数，则称为假分式。假分式通常必须先写成一个多项式与一个真分式之和，才能进行积分或拆分为部分分式。

In A-level papers, algebraic fractions rarely appear alone. They appear inside equations to be solved, inside expressions to be simplified before differentiation or integration, and inside partial fraction decompositions.

在 A-level 试卷中，代数分式很少单独出现，它常出现在待解的方程中、求导或积分前需化简的表达式中，以及部分分式分解的题目里。

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2. Domain Restrictions and Excluded Values | 定义域限制与排除值

r example, in (x² − 3x + 2)/(x − 1), the value x = 1 makes the denominator zero, so x ≠ 1.

在化简之前，应先写出需要排除的值。例如，在 (x² − 3x + 2)/(x − 1) 中，x = 1 会使分母为零，所以 x ≠ 1。

e. You must therefore state x ≠ 1 alongside x − 2, even though x = 1 would not make the simplified denominator zero. The cancellation removes the factor, not the restriction.

约分后得到 x − 2，但该化简式在 x = 1 处仍然无定义，因为原分式在这一点就没有定义。因此，在写出 x − 2 的同时必须注明 x ≠ 1，尽管 x = 1 并不会使化简后的分母为零。约分消去了因式，却没有消去限制条件。

The safe procedure is always the same: factorise the original denominator completely, set each factor equal to zero, and collect every value obtained. If the original denominator contains a factor that later cancels, that value is still excluded.

稳妥的做法始终一致：先把原分母完全因式分解，令每个因式等于零，收集所有得到的值。如果原分母中含有后来被约去的因式，该值依然要排除。

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3. Simplifying by Factorising | 因式分解化简

To simplify, factorise numerator and denominator completely, then cancel common polynomial factors. Example:

化简时，应先将分子和分母完全因式分解，再约去共同的整式因子。例如：

(x² + 5x + 6)/(x² − 4) = ((x + 2)(x + 3))/((x + 2)(x − 2)) = (x + 3)/(x − 2), x ≠ −2, 2

Notice that the factor x + 2 is cancelled, but the restriction x ≠ −2 must remain because it comes from the original denominator.

注意 x + 2 被约去，但限制条件 x ≠ −2 必须保留，因为它来自原分母。

+ bx + c = (x + p)(x + q) with p + q = b and pq = c, and grouping for four-term expressions. For quadratics with a leading coefficient other than 1, split the middle term or use the ac method.

最常用的因式分解包括：提公因式 ax + ay = a(x + y)、平方差 x² − a² = (x − a)(x + a)、二次三项式 x² + bx + c = (x + p)(x + q)（其中 p + q = b，pq = c），以及针对四项式的分组分解。对于首项系数不是 1 的二次式，可用拆中间项法或 ac 法。

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When the numerator and denominator share no common factor at all, the fraction is already in lowest terms and no further simplification is possible. Do not force a cancellation that is not there.

若分子与分母完全没有公因式，该分式已是最简形式，无法再约分。不要强行制造并不存在的约分。

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4. The Cancelling Trap: Terms Versus Factors | 约分陷阱：项与因式

Do not cancel individual terms inside a sum. For instance, (x + 2)/(x + 3) cannot be simplified to 2/3, and (x² + 1)/(x + 1) cannot be reduced by removing x from both numerator and denominator.

不要约去和式中的单个项。例如 (x + 2)/(x + 3) 不能化简为 2/3，(x² + 1)/(x + 1) 也不能通过消去分子和分母中的 x 来化简。

The reason is structural: cancellation is division by a common factor, and division does not distribute over addition or subtraction. You may only cancel objects that multiply the whole numerator and the whole denominator, never objects that are merely added into them.

原因在于结构：约分是除以一个公因式，而除法不能分配到加法或减法上。只有当某个对象整体乘以分子与分母时才能约去，仅仅是相加进去的部分绝对不能约去。

rue value is 3/4, not 2/3. Testing the original expression against the proposed simplification is one of the fastest ways to catch this mistake under exam pressure.

代入数值即可快速验证。令 x = 1，则 (x + 2)/(x + 3) 的真实值为 3/4，而不是 2/3。用原式检验化简结果，是考场上发现此类错误最快的方法之一。

Expression | 表达式	Valid step | 正确做法	Invalid step | 错误做法
(x² + 5x + 6)/(x² − 4)	Factorise, then cancel (x + 2)	Cancel x² and x²
(x + 2)/(x + 3)	Already in lowest terms	Cancel x to get 2/3
(2x + 4)/(2x + 6)	2(x + 2)/2(x + 3) = (x + 2)/(x + 3)	Cancel 2x to get 4/6

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5. Multiplying and Dividing | 乘法与除法

For multiplication, factorise all numerators and denominators first, cancel common factors, then multiply across. For division, multiply by the reciprocal of the divisor.

乘法运算时，先将所有分子和分母因式分解，约去公因式，然后相乘；除法运算则乘以除式的倒数。

/(x + 4) × (x + 4)/(x − 3) = ((x − 3)(x + 3))/(x + 4) × (x + 4)/(x − 3) = x + 3, x ≠ −4, 3

Here the factors (x + 4) and (x − 3) both cancel, leaving x + 3 for all allowed values of x. The excluded values −4 and 3 come from the original denominators and must be stated in the answer.

此处因式 (x + 4) 与 (x − 3) 都被约去，在所有允许的 x 取值下结果为 x + 3。排除值 −4 与 3 来自原分母，必须在答案中注明。

s (x² − 4)/(x + 1) × (x² − 1)/(x + 2), which factorises to ((x − 2)(x + 2)(x − 1)(x + 1))/((x + 1)(x + 2)) and simplifies to (x − 2)(x − 1) with x ≠ −1, −2, 1.

做除法时要注意，颠倒除式后它的分子会变成分母。表达式 (x² − 4)/(x + 1) ÷ (x + 2)/(x² − 1) 化为 (x² − 4)/(x + 1) × (x² − 1)/(x + 2)，因式分解后为 ((x − 2)(x + 2)(x − 1)(x + 1))/((x + 1)(x + 2))，化简得 (x − 2)(x − 1)，且 x ≠ −1, −2, 1。

Write the division as a single product before you cancel anything. Cancelling term by term across a division sign is one of the most common sources of lost marks in algebra questions.

在约分之前，先把除法整体写成一个乘积。跨越除号逐项约分，是代数题中最常见的失分来源之一。

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6. Adding and Subtracting | 加法与减法

Addition and subtraction require a common denominator. Factorise each denominator first, then take the lowest common multiple of the factors rather than the product of everything, so that the final answer stays simple.

加减法需要通分。先分解各分母，再取各因式的最小公倍式作为公分母，而不是把所有分母相乘，这样最终结果才足够简洁。

2/(x − 2) = (3(x − 2) + 2(x + 1))/((x + 1)(x − 2)) = (5x − 4)/((x + 1)(x − 2)), x ≠ −1, 2

When one denominator is a factor of the other, the common denominator is simply the larger one. For example, 1/(x² − 4) + 1/(x + 2) becomes 1/((x − 2)(x + 2)) + (x − 2)/((x − 2)(x + 2)), which combines to (x − 1)/((x − 2)(x + 2)).

若一个分母是另一个分母的因式，公分母直接取较大的那个即可。例如 1/(x² − 4) + 1/(x + 2) 变为 1/((x − 2)(x + 2)) + (x − 2)/((x − 2)(x + 2))，合并后得 (x − 1)/((x − 2)(x + 2))。

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