📚 IGCSE Mathematics Revision: Algebraic Fractions & Solving Equations | IGCSE 数学复习：代数分式与方程求解
Algebraic fractions are a core topic in IGCSE Mathematics. They combine fraction manipulation with algebraic techniques, and they often appear in both Paper 2 and Paper 4. Mastering them is essential for solving equations, simplifying expressions, and tackling word problems with confidence.
代数分式是 IGCSE 数学的核心内容。它把分数运算与代数技巧结合在一起，经常出现在 Paper 2 和 Paper 4 中。掌握代数分式，对于解方程、化简表达式以及自信地解决应用题都至关重要。
1. What Are Algebraic Fractions? | 什么是代数分式？
An algebraic fraction is a fraction where the numerator or the denominator (or both) contains an algebraic expression. For example:
代数分式是指分子或分母（或两者）中含有代数表达式的分数。例如：
3⁄x , (x + 2)⁄(x − 1) , 2x²⁄(x² + 1)
These are all algebraic fractions. They behave like ordinary fractions, but we must also consider values of the variable that would make the denominator zero.
这些都是代数分式。它们的运算规则与普通分数相同，但我们必须注意使分母为零的变量取值情况。
2. Simplifying Algebraic Fractions | 化简代数分式
To simplify an algebraic fraction, factorise the numerator and denominator completely, then cancel any common factors.
化简代数分式的方法是：将分子和分母完全因式分解，然后约去所有公因式。
Example 1 | 例 1
Simplify 6x²⁄(3x).
化简 6x²⁄(3x)。
6x²⁄(3x) = (6 × x × x)⁄(3 × x) = 2x
Example 2 | 例 2
Simplify (x² − 9)⁄(x² + 3x).
化简 (x² − 9)⁄(x² + 3x)。
Factorise the top and bottom:
对分子和分母分别因式分解：
(x² − 9)⁄(x² + 3x) = (x − 3)(x + 3)⁄(x)(x + 3) = (x − 3)⁄x
Remember that we can only cancel factors, not terms. For example, (x + 3) cancels because it is a common factor of both numerator and denominator.
记住，我们只能约去因式，而不能约去项。例如，(x + 3) 可以约去，因为它是分子和分母的公因式。
3. Multiplying and Dividing Algebraic Fractions | 代数分式的乘法和除法
Multiplication and division of algebraic fractions follow the same rules as numerical fractions.
代数分式的乘法和除法遵循与数值分数相同的规则。
Multiplication | 乘法
Multiply the numerators together and multiply the denominators together. Factorise first to simplify.
将分子与分子相乘，分母与分母相乘。先因式分解再化简。
(x + 1)⁄(x − 2) × (x − 2)²⁄(x² − 1)
Factorise where possible:
尽可能因式分解：
(x + 1)⁄(x − 2) × (x − 2)(x − 2)⁄(x − 1)(x + 1)
Cancel common factors:
约去公因式：
= (x − 2)⁄(x − 1)
Division | 除法
To divide by a fraction, multiply by its reciprocal.
除以一个分数，等于乘以它的倒数。
(x² + 5x + 6)⁄(x² − 4) ÷ (x + 3)⁄(x − 2)
Factorise all expressions:
对所有表达式因式分解：
= (x + 2)(x + 3)⁄(x − 2)(x + 2) × (x − 2)⁄(x + 3)
Cancel common factors:
约去公因式：
= 1
Always check whether the answer simplifies further.
始终检查答案是否还能进一步化简。
4. Adding and Subtracting Algebraic Fractions | 代数分式的加法和减法
To add or subtract algebraic fractions, first find a common denominator — usually the least common multiple (LCM) of the denominators.
进行代数分式的加法或减法时，首先要找到公分母——通常是各分母的最小公倍数 (LCM)。
Example 1 | 例 1
Simplify 2⁄x + 3⁄(x + 1).
化简 2⁄x + 3⁄(x + 1)。
The LCM is x(x + 1).
最小公倍数为 x(x + 1)。
2⁄x + 3⁄(x + 1) = 2(x + 1)⁄(x(x + 1)) + 3x⁄(x(x + 1))
= (2x + 2 + 3x)⁄(x(x + 1)) = (5x + 2)⁄(x(x + 1))
Example 2 | 例 2
Simplify 1⁄(x − 2) − 1⁄(x + 2).
化简 1⁄(x − 2) − 1⁄(x + 2)。
= (x + 2)⁄((x − 2)(x + 2)) − (x − 2)⁄((x − 2)(x + 2))
= (x + 2 − x + 2)⁄((x − 2)(x + 2)) = 4⁄(x² − 4)
Be careful with signs when subtracting the second numerator.
减去第二个分子时，要特别注意符号。
5. Solving Equations with Algebraic Fractions | 求解含代数分式的方程
When solving equations involving algebraic fractions, multiply every term by the common denominator to eliminate the fractions, then solve the resulting equation.
求解含代数分式的方程时，将每一项都乘以公分母，从而消去分数，然后求解所得方程。
Example | 例
Solve 3⁄x + 2 = 5⁄(x + 1).
解方程 3⁄x + 2 = 5⁄(x + 1)。
Multiply both sides by x(x + 1):
两边同时乘以 x(x + 1)：
3(x + 1) + 2x(x + 1) = 5x
Expand and simplify:
展开并化简：
3x + 3 + 2x² + 2x = 5x
2x² + 5x + 3 = 5x
2x² + 3 = 0
This has no real solution because x² cannot be negative. Always check that your solutions do not make any denominator zero.
该方程无实数解，因为 x² 不可能为负数。务必检查求得的解是否会使任何一个分母为零。
6. Extraneous Solutions | 增根
When multiplying both sides of an equation by a variable expression, you may introduce extraneous solutions — values that satisfy the transformed equation but not the original one.
当方程两边同时乘以一个含变量的表达式时，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。
Example | 例
Solve 2x⁄(x − 1) = 4⁄(x − 1) + 1.
解方程 2x⁄(x − 1) = 4⁄(x − 1) + 1。
Multiply by (x − 1):
两边乘以 (x − 1)：
2x = 4 + x − 1
2x = x + 3
x = 3
Check: when x = 3, the denominator is 2, not zero, so it is a valid solution.
检验：当 x = 3 时，分母为 2，不为零，所以它是有效解。
1 had appeared as a solution, it would be extraneous because the denominator becomes zero.
如果 x = 1 作为解出现，那么它就是增根，因为此时分母为零。
7. Common Mistakes and How to Avoid Them | 常见错误与规避方法
Mistake: Cancelling terms instead of factors. Only common factors can be cancelled.
错误：约去项而不是约去因式。只有公因式才能约去。
Mistake: Forgetting to check for zero denominators. Always state that x cannot equal values that make a denominator zero.
错误：忘记检查分母为零的情况。始终要注明 x 不能等于使分母为零的值。
Mistake: Sign errors in subtraction. When subtracting a fraction, subtract the entire numerator.
错误：减法中的符号错误。减去一个分数时，要减去整个分子。
Mistake: Not factorising first. Always factorise before multiplying or cancelling to simplify your work.
错误：没有先因式分解。在乘法或约分之前，务必先因式分解，以简化运算。
8. Worked Exam-Style Question | 典型考试题解析
Question | 题目
Solve the equation 3⁄(x + 2) − 1⁄(x − 2) = 4⁄(x² − 4).
解方程 3⁄(x + 2) − 1⁄(x − 2) = 4⁄(x² − 4)。
Solution | 解答
Notice that x² − 4 = (x + 2)(x − 2). So the common denominator is (x + 2)(x − 2).
注意到 x² − 4 = (x + 2)(x − 2)。因此公分母为 (x + 2)(x − 2)。
3(x − 2) − 1(x + 2) = 4
Expand the numerators:
展开分子：
3x − 6 − x − 2 = 4
2x − 8 = 4
2x = 12
x = 6
heck: x = 6 gives denominators of 8 and 4, both non-zero. Therefore x = 6 is the solution.
检验：x = 6 时分母分别为 8 和 4，均不为零。因此 x = 6 是原方程的解。
9. Practice Questions | 练习题目
Simplify (x² + 4x + 3)⁄(x² + 2x + 1).
化简 (x² + 4x + 3)⁄(x² + 2x + 1)。
Simplify 2⁄(x² − 1) + 3⁄(x + 1).
化简 2⁄(x² − 1) + 3⁄(x + 1)。
Solve 4⁄(x − 3) = 2x⁄(x − 3) + 1.
解方程 4⁄(x − 3) = 2x⁄(x − 3) + 1。
Solve 1⁄(x + 1) + 2⁄(x − 1) = 1.
解方程 1⁄(x + 1) + 2⁄(x − 1) = 1。
Answers | 参考答案
- (x + 3)⁄(x + 1)
- (3x − 1)⁄(x² − 1)
- x = 3 is extraneous; no solution.
- x = 0 or x = 3 (check both).
10. Key Takeaways | 核心要点
Always factorise expressions before simplifying algebraic fractions.
化简代数分式前，一定要先进行因式分解。
Only cancel common factors, never terms.
只能约去公因式，绝不能约去项。
Find a common denominator before adding or subtracting fractions.
进行加减运算前，先找到公分母。
When solving equations, multiply through by the common denominator and check for extraneous solutions.
解方程时，整体乘以公分母，并检查是否有增根。
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