📚 Adding and Subtracting Algebraic Fractions | 代数分式的加减运算
In this revision guide, we will learn how to add and subtract algebraic fractions step by step. This skill is essential for solving equations and simplifying expressions in Edexcel IGCSE Mathematics.
在本复习指南中，我们将一步步学习如何对代数分式进行加减运算。这是 Edexcel IGCSE 数学中解方程和化简表达式的重要技能。
1. What Are Algebraic Fractions? | 什么是代数分式？
An algebraic fraction is a fraction whose numerator or denominator contains variables. For example, 3/x and (x+1)/(x-2) are algebraic fractions.
代数分式是指分子或分母中含有变量的分数。例如，3/x 和 (x+1)/(x-2) 都是代数分式。
Division by zero is undefined, so we must state restrictions. In 3/x, x cannot be 0; in (x+1)/(x-2), x cannot be 2.
由于不能除以零，我们必须说明限制条件。在 3/x 中，x 不能为 0；在 (x+1)/(x-2) 中，x 不能为 2。
2. Review: Factorisation | 复习：因式分解
Before adding or subtracting algebraic fractions, we often need to factorise denominators to find a common denominator. Here are the most common factorisation types you must know.
在进行代数分式的加减运算前，我们通常需要将分母因式分解，以找到公分母。以下是必须掌握的最常见因式分解类型。
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Common factor: ax + ay = a(x+y)
提取公因式：ax + ay = a(x+y)
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Quadratic trinomial: x² + 5x + 6 = (x+2)(x+3)
二次三项式：x² + 5x + 6 = (x+2)(x+3)
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Difference of squares: x² - 9 = (x-3)(x+3)
平方差：x² - 9 = (x-3)(x+3)
3. Adding and Subtracting Fractions with Same Denominator | 同分母分式的加减
When the denominators are the same, we simply add or subtract the numerators and keep the denominator unchanged. Then we simplify if possible.
当分母相同时，我们只需将分子相加或相减，分母保持不变，然后在可能的情况下进行化简。
2x/(x+1) + 3/(x+1) = (2x+3)/(x+1)
For subtraction, be careful to use brackets around the second numerator.
对于减法，注意在第二个分子上加上括号。
5x/(x-2) - (x+1)/(x-2) = (5x - (x+1))/(x-2) = (4x - 1)/(x-2)
4. Finding the Lowest Common Denominator (LCD) | 求最小公分母
To add or subtract fractions with different denominators, we first find the Lowest Common Denominator (LCD). The LCD is the product of all different factors in the denominators, using the highest power of each factor.
要对异分母分式进行加减，我们首先要找到最小公分母（LCD）。LCD 是各分母中所有不同因式的乘积，每个因式取最高次幂。
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For numerical denominators, use the least common multiple. For example, the LCD of 6 and 10 is 30.
对于数字分母，使用最小公倍数。例如，6 和 10 的 LCD 是 30。
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(x² + 2x + 1), since x² - 1 = (x-1)(x+1) and x² + 2x + 1 = (x+1)², the LCD is (x-1)(x+1)².
对于代数分母，先因式分解。例如，(x² - 1) = (x-1)(x+1)，(x² + 2x + 1) = (x+1)²，因此 LCD 是 (x-1)(x+1)²。
5. Adding and Subtracting Fractions with Different Denominators | 异分母分式的加减
Use this step-by-step method for any pair of algebraic fractions.
对于任何一对代数分式，使用以下步骤。
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Factorise each denominator completely.
将每个分母完全因式分解。
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Find the LCD of all denominators.
求出所有分母的 LCD。
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Rewrite each fraction with the LCD as its new denominator.
将每个分式改写为以 LCD 作为新分母的形式。
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Add or subtract the numerators, keeping the LCD.
保持分母为 LCD，将分子相加或相减。
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Simplify the result by cancelling common factors if possible.
如果可能，通过约去公因式来化简结果。
1/x + 1/(x+1) = ((x+1) + x)/(x(x+1)) = (2x+1)/(x(x+1))
6. Handling Denominators That Need Factorising | 处理需要因式分解的分母
Some denominators are not given in factored form. You must factorise them first to find the LCD.
有些分母并不是因式分解后的形式。你必须先将它们因式分解，才能找到 LCD。
Example: Simplify 2/(x² - 4) + 3/(x+2).
例如：化简 2/(x² - 4) + 3/(x+2)。
2)(x+2), the LCD is (x-2)(x+2). The second fraction needs to be multiplied by (x-2)/(x-2).
由于 x² - 4 = (x-2)(x+2)，LCD 是 (x-2)(x+2)。第二个分式需要乘以 (x-2)/(x-2)。
2/((x-2)(x+2)) + 3(x-2)/((x-2)(x+2)) = (2 + 3(x-2))/((x-2)(x+2)) = (3x - 4)/((x-2)(x+2))
7. Simplifying the Result | 化简结果
After adding or subtracting, always check whether the numerator and denominator share a common factor. If they do, cancel it.
在加减运算之后，始终检查分子和分母是否有公因式。如果有，就将其约去。
(x+1)/((x+1)(x-1)) = 1/(x-1), provided x ≠ -1
Remember that you cannot cancel terms, only factors. For example, in (x+1)/(x+2), the x cannot be cancelled because x is not a factor of the whole numerator.
记住，不能约去项，只能约去因式。例如，在 (x+1)/(x+2) 中，不能约去 x，因为 x 不是整个分子的因式。
8. Common Mistakes and Pitfalls | 常见错误与陷阱
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Forgetting brackets when subtracting numerators: 5x/(x-2) - (x+1)/(x-2) must be written as (5x - (x+1))/(x-2), not 5x - x + 1.
减法中忘记括号：5x/(x-2) - (x+1)/(x-2) 必须写成 (5x - (x+1))/(x-2)，而不是 5x - x + 1。
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Adding or subtracting denominators instead of keeping the common denominator.
将分母相加或相减，而没有保持公分母不变。
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Cancelling incorrectly, such as cancelling x from (x+1)/(x+2).
错误约分，例如在 (x+1)/(x+2) 中约去 x。
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Forgetting to factorise denominators before finding the LCD.
在求 LCD 之前忘记因式分解分母。
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Ignoring the zero-denominator restrictions when stating the final answer.
在写出最终答案时忽略分母不能为零的限制条件。
9. Worked Examples | 综合例题
Example 1: Simplify 2x/(x² - 9) + 1/(x - 3).
例题 1：化简 2x/(x² - 9) + 1/(x - 3)。
x² - 9 = (x-3)(x+3), so LCD = (x-3)(x+3)
+3)/((x-3)(x+3)) = (2x + x + 3)/((x-3)(x+3)) = (3x + 3)/((x-3)(x+3)) = 3(x+1)/((x-3)(x+3))
Example 2: Simplify 3/(x² + 2x) - 2/(x + 2).
例题 2：化简 3/(x² + 2x) - 2/(x + 2)。
x² + 2x = x(x+2), so LCD = x(x+2)
3/(x(x+2)) - 2x/(x(x+2)) = (3 - 2x)/(x(x+2))
10. Practice Questions | 练习题
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1. Simplify: 3/(x+1) + 2/(x+1).
1. 化简：3/(x+1) + 2/(x+1)。
Answer: 5/(x+1).
答案：5/(x+1)。
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2. Simplify: 1/(x-2) - 1/(x+3).
2. 化简：1/(x-2) - 1/(x+3)。
Answer: 5/((x-2)(x+3)).
答案：5/((x-2)(x+3))。
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3. Simplify: 2/(x² - 1) + 1/(x² + 2x + 1).
3. 化简：2/(x² - 1) + 1/(x² + 2x + 1)。
Answer: (3x + 1)/((x-1)(x+1)²).
答案：(3x + 1)/((x-1)(x+1)²)。
11. Summary | 总结
To add or subtract algebraic fractions, always factorise denominators first, find the LCD, rewrite each fraction with that common denominator, then combine the numerators and simplify. Remember to state any restrictions on the variable.
要进行代数分式的加减运算，务必先因式分解分母，找到 LCD，将每个分式改写为以该公分母为分母的形式，然后合并分子并化简。同时记得写出变量的限制条件。
With regular practice, this skill becomes fast and reliable. Make sure you master the factorisation techniques and always check your final answer.
通过经常练习，这项技能会变得快速而可靠。请务必掌握因式分解技巧，并始终检查你的最终答案。
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