Algebraic Fractions: Simplification, Operations and Equation Solving | 代数分式：化简、运算与方程求解

📚 Algebraic Fractions: Simplification, Operations and Equation Solving | 代数分式：化简、运算与方程求解

Algebraic fractions are fractions in which the numerator and/or denominator are algebraic expressions. Mastering them is essential for success in A-Level Mathematics, as they appear throughout pure mathematics, from solving equations to calculus and beyond.

代数分式是指分子和（或）分母中包含代数表达式的分数。掌握代数分式的运算是A-Level数学成功的关键，因为它们在纯数学中无处不在——从解方程到微积分乃至更高级的内容。

1. Understanding Algebraic Fractions | 理解代数分式

An algebraic fraction is a fraction where the numerator, the denominator, or both are algebraic expressions. Examples include ³⁄ₓ, (x² + 1)⁄(x − 3), and (2x + 5)⁄(x² − 4).

代数分式是指分子、分母或两者均为代数表达式的分数。例如 ³⁄ₓ、(x² + 1)⁄(x − 3) 和 (2x + 5)⁄(x² − 4)。

Just like numerical fractions, algebraic fractions follow the same fundamental rules of arithmetic. However, because they involve variables, we must pay careful attention to the domain — the values of the variable for which the fraction is defined.

与数值分数一样，代数分式遵循相同的基本运算规则。然而，由于它们包含变量，我们必须特别注意定义域——即分式有意义的变量取值范围。

For a fraction a⁄b to be defined, b cannot equal zero. For example, in the fraction 5⁄(x − 2), the denominator becomes zero when x = 2, so we say the fraction is undefined at x = 2.

对于分式 a⁄b，分母 b 不能等于零。例如，在分式 5⁄(x − 2) 中，当 x = 2 时分母为零，因此我们说该分式在 x = 2 处无定义。

Domain restriction: b ≠ 0 | 定义域限制：b ≠ 0

In x⁄(x² − 1), the denominator factors as (x − 1)(x + 1), so x ≠ 1 and x ≠ −1.
在 x⁄(x² − 1) 中，分母因式分解为 (x − 1)(x + 1)，因此 x ≠ 1 且 x ≠ −1。
In 2x⁄(x² + 4), since x² + 4 ≥ 4 always, the denominator never equals zero, so no restrictions apply.
在 2x⁄(x² + 4) 中，因为 x² + 4 ≥ 4 恒成立，分母永不为零，所以没有限制条件。

Understanding the domain of an algebraic fraction is not merely a technical detail — it directly affects the validity of any simplification or solution. When we cancel common factors, we must note any excluded values.

理解代数分式的定义域不仅仅是一个技术细节——它直接关系到任何化简或解的有效性。当我们约去公因式时，必须注意并记录被排除的值。

2. Simplifying Algebraic Fractions | 化简代数分式

The first step in working with algebraic fractions is simplification. A fraction is in its simplest form when the numerator and denominator share no common algebraic factors.

处理代数分式的第一步是化简。当分子与分母没有共同的代数因式时，分式即为最简形式。

To simplify an algebraic fraction, we begin by factoring both the numerator and the denominator completely. Then we cancel any common factors that appear in both.

化简代数分式时，我们首先将分子和分母完全因式分解，然后约去两者共有的因式。

Example: Simplify (x² − 9)⁄(x² − 3x).

例如：化简 (x² − 9)⁄(x² − 3x)。

Step 1 — Factor the numerator: x² − 9 = (x − 3)(x + 3).

第一步——分解分子：x² − 9 = (x − 3)(x + 3)。

Step 2 — Factor the denominator: x² − 3x = x(x − 3).

第二步——分解分母：x² − 3x = x(x − 3)。

Step 3 — Cancel the common factor (x − 3):

第三步——约去公因式 (x − 3)：

(x² − 9)⁄(x² − 3x) = (x − 3)(x + 3)⁄x(x − 3) = (x + 3)⁄x

The simplified form is (x + 3)⁄x, with the restriction x ≠ 0 and x ≠ 3 because the original denominator cannot be zero.

化简结果为 (x + 3)⁄x，限制条件为 x ≠ 0 且 x ≠ 3，因为原分母不能为零。

Always factor completely before cancelling — you cannot cancel terms that are added, only factors that are multiplied.
在约分之前必须完全因式分解——不能约去相加的项，只能约去相乘的因式。
Note all restrictions on the variable from the original expression.
记下原始表达式中对变量的所有限制条件。

3. Multiplying Algebraic Fractions | 代数分式的乘法

Multiplying algebraic fractions follows the same rule as multiplying numerical fractions: multiply the numerators together and multiply the denominators together.

代数分式的乘法与数值分数的乘法规则相同：分子相乘得分子，分母相乘得分母。

Procedure: Multiply Numerators and Multiply Denominators (简化前先交叉约分).

步骤：分子乘分子、分母乘分母（可在相乘前先进行交叉约分）。

It is often easier to factor and cancel before performing the multiplication, as this reduces the size of the numbers and expressions involved.

通常先因式分解并约分再相乘会更简便，因为这样可以减小涉及的数值和表达式的规模。

Example: Simplify (x² − 1)⁄(x² − 4) × (x − 2)⁄(x + 1).

例如：化简 (x² − 1)⁄(x² − 4) × (x − 2)⁄(x + 1)。

First, factor everything:

首先，将所有表达式因式分解：

(x − 1)(x + 1)⁄(x − 2)(x + 2) × (x − 2)⁄(x + 1)

Now cancel common factors diagonally: (x + 1) cancels with (x + 1), and (x − 2) cancels with (x − 2).

现在进行对角线约分：(x + 1) 与 (x + 1) 约去，(x − 2) 与 (x − 2) 约去。

= (x − 1)⁄(x + 2)

The answer is (x − 1)⁄(x + 2), valid for x ≠ ±2 and x ≠ −1.

答案为 (x − 1)⁄(x + 2)，在 x ≠ ±2 且 x ≠ −1 的条件下有意义。

4. Dividing Algebraic Fractions | 代数分式的除法

Division of algebraic fractions is performed by multiplying by the reciprocal (also called the multiplicative inverse) of the divisor.

代数分式的除法通过乘以除数的倒数（也称为乘法逆元）来进行。

This means we keep the first fraction exactly as it is, flip the second fraction upside down, and then multiply. After flipping, we follow all the rules of multiplication described above.

这意味着保持第一个分式不变，将第二个分式上下颠倒，然后进行乘法运算。翻转后，我们按照上述乘法规则进行计算。

Example: Simplify (x² − 4)⁄(x + 3) ÷ (x − 2)⁄(x² − 9).

例如：化简 (x² − 4)⁄(x + 3) ÷ (x − 2)⁄(x² − 9)。

Step 1 — Flip the second fraction:

第一步——翻转第二个分式：

(x² − 4)⁄(x + 3) × (x² − 9)⁄(x − 2)

Step 2 — Factor everything:

第二步——全部因式分解：

(x − 2)(x + 2)⁄(x + 3) × (x − 3)(x + 3)⁄(x − 2)

Step 3 — Cancel common factors: (x − 2) and (x + 3):

第三步——约去公因式 (x − 2) 和 (x + 3)：

= (x + 2)(x − 3) = x² − x − 6

Note the restrictions: x ≠ ±3, x ≠ 2, and also x ≠ −3 from the original divisor’s denominator.

注意限制条件：x ≠ ±3、x ≠ 2，且原除数分母要求 x ≠ −3。

5. Adding and Subtracting Algebraic Fractions | 代数分式的加减法

Adding and subtracting algebraic fractions requires a common denominator, exactly as with numerical fractions. The most efficient common denominator is the lowest common multiple (LCM) of the denominators.

与数值分数一样，代数分式的加减需要公分母。最有效的公分母是各分母的最小公倍数（LCM）。

Step 1 — Factor each denominator fully.

第一步——将每个分母完全因式分解。

Step 2 — Determine the LCM of the denominators.

第二步——求出分母的最小公倍数。

Step 3 — Rewrite each fraction with the common denominator and perform the addition or subtraction on the numerators only.

第三步——将每个分式转化为以公分母为分母的形式，然后仅在分子上进行加减运算。

Step 4 — Simplify the resulting fraction if possible.

第四步——尽可能化简结果分式。

Example: Simplify 3⁄(x − 2) + 2⁄(x + 3).

例如：化简 3⁄(x − 2) + 2⁄(x + 3)。

The denominators are already in factored form. The LCM is (x − 2)(x + 3).

分母已经是因式分解形式。最小公倍数为 (x − 2)(x + 3)。

Rewrite each fraction with the common denominator:

将每个分式转化为以公分母为分母的形式：

= 3(x + 3)⁄[(x − 2)(x + 3)] + 2(x − 2)⁄[(x − 2)(x + 3)]

Now combine the numerators:

现在合并分子：

= [3(x + 3) + 2(x − 2)]⁄[(x − 2)(x + 3)] = (3x + 9 + 2x − 4)⁄[(x − 2)(x + 3)]

= (5x + 5)⁄[(x − 2)(x + 3)] = 5(x + 1)⁄[(x − 2)(x + 3)]

The final answer is 5(x + 1)⁄[(x − 2)(x + 3)], valid for x ≠ 2 and x ≠ −3.

最终答案为 5(x + 1)⁄[(x − 2)(x + 3)]，在 x ≠ 2 且 x ≠ −3 的条件下有意义。

6. Working with More Complex Denominators | 处理更复杂的分母

Often the denominators in sum or difference problems are not given in factored form. In such cases, factoring each denominator first is essential.

在加减问题中，分母往往不是因式分解形式。在这种情况下，首先分解每个分母是至关重要的。

Consider the example: 2⁄(x² − 1) − 1⁄(x² + 2x + 1).

考虑以下例子：2⁄(x² − 1) − 1⁄(x² + 2x + 1)。

Factor each denominator:

分解每个分母：

x² − 1 = (x − 1)(x + 1) and x² + 2x + 1 = (x + 1)²

The LCM of (x − 1)(x + 1) and (x + 1)² is (x − 1)(x + 1)².

(x − 1)(x + 1) 与 (x + 1)² 的最小公倍数为 (x − 1)(x + 1)²。

Rewrite each fraction:

重写每个分式：

= 2(x + 1)⁄[(x − 1)(x + 1)²] − (x − 1)⁄[(x − 1)(x + 1)²]

Combine the numerators:

合并分子：

= [2(x + 1) − (x − 1)]⁄[(x − 1)(x + 1)²] = (2x + 2 − x + 1)⁄[(x − 1)(x + 1)²]

= (x + 3)⁄[(x − 1)(x + 1)²]

The result is (x + 3)⁄[(x − 1)(x + 1)²], with x ≠ ±1.

结果为 (x + 3)⁄[(x − 1)(x + 1)²]，其中 x ≠ ±1。

7. Partial Fractions: The Reverse Process | 部分分式：反向操作

Partial fractions involve expressing a single complicated algebraic fraction as the sum of two or more simpler fractions. This technique is particularly useful in integration and in solving differential equations.

部分分式是将一个复杂代数分式表示为两个或更多简单分式之和。这一技巧在积分和求解微分方程中特别有用。

We use partial fractions when the degree of the numerator is less than the degree of the denominator and the denominator can be factored into distinct linear factors, repeated linear factors, or irreducible quadratic factors.

当分子的次数低于分母的次数，且分母可以分解为不同的线性因式、重复的线性因式或不可约的二次因式时，我们使用部分分式。

Example: Express 5x − 1⁄[(x − 2)(x + 1)] as partial fractions.

例如：将 5x − 1⁄[(x − 2)(x + 1)] 表示为部分分式。

We set up the form:

我们设定以下形式：

5x − 1⁄[(x − 2)(x + 1)] = A⁄(x − 2) + B⁄(x + 1)

Multiply both sides by (x − 2)(x + 1):

两边同时乘以 (x − 2)(x + 1)：

5x − 1 = A(x + 1) + B(x − 2)

Now choose suitable values of x to find A and B.

现在选择适当的 x 值来求 A 和 B。

Let x = 2: 5(2) − 1 = A(3) ⇒ 9 = 3A ⇒ A = 3.

令 x = 2：5(2) − 1 = A(3) ⇒ 9 = 3A ⇒ A = 3。

Let x = −1: 5(−1) − 1 = B(−3) ⇒ −6 = −3B ⇒ B = 2.

令 x = −1：5(−1) − 1 = B(−3) ⇒ −6 = −3B ⇒ B = 2。

Therefore:

因此：

5x − 1⁄[(x − 2)(x + 1)] = 3⁄(x − 2) + 2⁄(x + 1)

This decomposition into partial fractions can then be readily integrated or manipulated in later work.

这种部分分式分解在后续的积分或运算中可以直接使用。

8. Solving Equations Involving Algebraic Fractions | 求解含代数分式的方程

Equations involving algebraic fractions are solved by eliminating the fractions first. The standard method is to multiply both sides of the equation by the LCM of all denominators present.

解含代数分式的方程时，首先要消去分式。标准方法是方程两边同时乘以所有分母的最小公倍数。

However, this process can introduce extraneous solutions — values that satisfy the transformed equation but not the original equation because they make a denominator zero. Therefore, checking your final answers is mandatory.

然而，这一过程可能会引入增根——即满足变换后方程但不满足原方程的数值，因为它们使某个分母为零。因此，必须检验最终答案。

Example: Solve 2⁄(x − 1) + x⁄(x + 2) = 3.

例如：解方程 2⁄(x − 1) + x⁄(x + 2) = 3。

The LCM of (x − 1) and (x + 2) is (x − 1)(x + 2). Multiply both sides:

(x − 1) 与 (x + 2) 的最小公倍数为 (x − 1)(x + 2)。两边同时乘以：

2(x + 2) + x(x − 1) = 3(x − 1)(x + 2)

Expand both sides:

展开两边：

2x + 4 + x² − x = 3(x² + x − 2)

x² + x + 4 = 3x² + 3x − 6

Rearrange to obtain a quadratic equation:

整理得到二次方程：

0 = 2x² + 2x − 10 ⇒ x² + x − 5 = 0

Using the quadratic formula:

使用求根公式：

x = [−1 ± √(1 + 20)]⁄2 = (−1 ± √21)⁄2

Check: Neither x = (−1 + √21)⁄2 nor x = (−1 − √21)⁄2 equals 1 or −2, so both are valid.

检验：x = (−1 + √21)⁄2 和 x = (−1 − √21)⁄2 都不等于 1 或 −2，因此两个解均有效。

9. Word Problems and Applications | 应用题与实际应用

Algebraic fractions frequently arise in applied problems involving rates, proportions, and mixtures. Translating a word problem into an equation with algebraic fractions is a key skill.

代数分式经常出现在涉及速率、比例和混合问题的实际应用中。将文字题转化为含代数分式的方程是一项关键技能。

Consider the classic rate problem: A pipe can fill a tank in x hours, while a second pipe can fill the same tank in x − 3 hours. Together, they fill the tank in 2 hours. Find x.

考虑一个经典的速率问题：一根水管注满一个水箱需要 x 小时，而第二根水管注满同样水箱需要 x − 3 小时。两根水管一起注满需要 2 小时。求 x。

The rates are 1⁄x and 1⁄(x − 3) tankfuls per hour. Working together:

两根水管的速率分别为 1⁄x 和 1⁄(x − 3) 水箱/小时。一起工作时：

1⁄x + 1⁄(x − 3) = 1⁄2

Multiply both sides by 2x(x − 3):

两边同时乘以 2x(x − 3)：

2(x − 3) + 2x = x(x − 3)

2x − 6 + 2x = x² − 3x ⇒ 4x − 6 = x² − 3x

x² − 7x + 6 = 0 ⇒ (x − 1)(x − 6) = 0

Thus x = 1 or x = 6. However, if x = 1, then x − 3 = −2, which is impossible for a time value. The valid solution is x = 6.

因此 x = 1 或 x = 6。但如果 x = 1，则 x − 3 = −2，作为时间值不可能。有效解为 x = 6。

This example illustrates the importance of interpreting solutions in context. A mathematically valid value may not be a physically meaningful answer.

这个例子说明了结合具体背景解释解的重要意义。数学上有效的解不一定在实际物理情境中有意义。

10. Common Mistakes and Pitfalls | 常见错误与易错点

Students often encounter several recurring difficulties when working with algebraic fractions. Recognising these common errors helps avoid them.

学生在学习代数分式时经常遇到一些反复出现的困难。识别这些常见错误有助于避免它们。

One of the most frequent mistakes is cancelling terms that are not factors. For example, simplifying (x + 3)⁄(x + 5) by cancelling the x’s to get 3⁄5 is incorrect because x is a term, not a factor.

最常见的错误之一是约去并非因式的项。例如，将 (x + 3)⁄(x + 5) 中的 x 约去得到 3⁄5 是错误的，因为 x 是项，不是因式。

Incorrect \| 错误	Correct \| 正确	Reason \| 原因
(x + 3)⁄(x + 5) = 3⁄5	Cannot simplify	x is a term, not a factor
(x² + 4)⁄(x + 2) = x + 2	Cannot simplify	Numerator does not factor as (x + 2)

Another common error is forgetting to check for extraneous solutions when solving equations. Always verify that your solutions do not make any denominator zero in the original equation.

另一个常见错误是在解方程时忘记检验增根。务必确认你的解不会使原方程中的任何分母为零。

When subtracting fractions, the error of misapplying the minus sign is also common. It is essential to apply the subtraction to the entire numerator, not just the first term.

在分数相减时，错误地处理负号也很常见。必须将减号应用于整个分子，而不只是第一项。

For example, in 3⁄(x + 1) − (2x + 1)⁄(x + 1), the correct numerator after combining is 3 − (2x + 1) = 2 − 2x, not 3 − 2x + 1.

例如，在 3⁄(x + 1) − (2x + 1)⁄(x + 1) 中，合并后的正确分子是 3 − (2x + 1) = 2 − 2x，而不是 3 − 2x + 1。

11. Practice Problems | 练习题目

To consolidate the techniques covered in this article, attempt the following problems without looking at the solutions first. These represent the standard level expected in A-Level examinations.

为了巩固本文所学的技巧，请先尝试解答以下问题，再看答案。这些题目的难度对应A-Level考试的常规水平。

Problem 1: Simplify (x² − 16)⁄(x² − 4x).

题目1：化简 (x² − 16)⁄(x² − 4x)。

Problem 2: Simplify 2⁄(x + 1) − 1⁄(x − 2).

题目2：化简 2⁄(x + 1) − 1⁄(x − 2)。

Problem 3: Solve 3⁄(x + 2) + 1⁄(x − 1) = 2.

题目3：解方程 3⁄(x + 2) + 1⁄(x − 1) = 2。

Problem 4: Express (3x + 1)⁄[(x − 1)(x + 2)] as partial fractions.

题目4：将 (3x + 1)⁄[(x − 1)(x + 2)] 表示为部分分式。

Solution 1: Factor to get (x − 4)(x + 4)⁄x(x − 4) = (x + 4)⁄x, x ≠ 0, 4.

答案1：分解得 (x − 4)(x + 4)⁄x(x − 4) = (x + 4)⁄x，x ≠ 0, 4。

Solution 2: The common denominator is (x + 1)(x − 2): [2(x − 2) − (x + 1)]⁄[(x + 1)(x − 2)] = (x − 5)⁄[(x + 1)(x − 2)], x ≠ −1, 2.

答案2：公分母为 (x + 1)(x − 2)：[2(x − 2) − (x + 1)]⁄[(x + 1)(x − 2)] = (x − 5)⁄[(x + 1)(x − 2)]，x ≠ −1, 2。

Solution 3: Multiply by (x + 2)(x − 1): 3(x − 1) + (x + 2) = 2(x + 2)(x − 1). This simplifies to 3x − 3 + x + 2 = 2x² + 2x − 4, i.e. 2x² − 2x − 3 = 0. Therefore x = (2 ± √28)⁄4 = (1 ± √7)⁄2, both valid.

答案3：两边乘以 (x + 2)(x − 1)：3(x − 1) + (x + 2) = 2(x + 2)(x − 1)。化简得 3x − 3 + x + 2 = 2x² + 2x − 4，即 2x² − 2x − 3 = 0。因此 x = (2 ± √28)⁄4 = (1 ± √7)⁄2，两个解均有效。

Solution 4: Set (3x + 1)⁄[(x − 1)(x + 2)] = A⁄(x − 1) + B⁄(x + 2). Then 3x + 1 = A(x + 2) + B(x − 1). When x = 1: 4 = 3A ⇒ A = 4⁄3. When x = −2: −5 = −3B ⇒ B = 5⁄3. Hence the result is 4⁄[3(x − 1)] + 5⁄[3(x + 2)].

答案4：设 (3x + 1)⁄[(x − 1)(x + 2)] = A⁄(x − 1) + B⁄(x + 2)。则 3x + 1 = A(x + 2) + B(x − 1)。当 x = 1：4 = 3A ⇒ A = 4⁄3。当 x = −2：−5 = −3B ⇒ B = 5⁄3。因此结果为 4⁄[3(x − 1)] + 5⁄[3(x + 2)]。

12. Summary and Key Takeaways | 总结与核心要点

Algebraic fractions are a foundational topic in A-Level mathematics. Mastery of their simplification, arithmetic operations, and use in equations is essential for solving both exam problems and more advanced mathematical problems.

代数分式是A-Level数学中的基础性话题。熟练掌握它们的化简、运算以及在方程中的运用，对于解答考试题目和更高级的数学问题都至关重要。

The most important ideas to remember are: always factor completely before simplifying; when adding or subtracting, use the LCM of the denominators; when dividing, multiply by the reciprocal; and always verify solutions to equations by checking the domain of the original fractions.

需要记住的最重要的概念是：化简前务必完全因式分解；加减时要使用分母的最小公倍数；除法要乘以倒数；解方程后一定要通过原分式的定义域来验根。

Factor first, then cancel — never cancel terms directly.
先因式分解再约分——绝不能直接约去项。
The LCM of denominators handles addition and subtraction efficiently.
利用分母的最小公倍数可以高效地进行加减法。
Multiply by the reciprocal when dividing fractions.
除以一个分式等于乘以它的倒数。
Always state domain restrictions and check for extraneous solutions.
始终注明定义域限制并检查是否产生增根。
Practise regularly — proficiency comes from solving many varied problems.
定期练习——熟练来自于解决大量不同类型的题目。

With consistent practice, solving algebraic fraction problems becomes an automatic and reliable process, providing a solid foundation for everything from integration to differential equations in your continued studies.

通过持续练习，解决代数分式问题将成为一种自动而可靠的技能，为你后续学习中从积分到微分方程的所有内容奠定坚实基础。

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Algebraic Fractions: Simplification, Operations and Equation Solving | 代数分式：化简、运算与方程求解