Algebraic Fractions | 代数分式

📚 Algebraic Fractions | 代数分式

Algebraic fractions are those where the numerator, denominator, or both contain algebraic expressions, such as (x+2)/(x-3). In IGCSE Edexcel Mathematics, mastering algebraic fractions is essential for simplifying expressions, solving equations, and progressing to more advanced topics like functions and calculus. This article provides a clear, step-by-step guide to algebraic fractions, from basic simplification to solving complex fractional equations.

代数分式是指分子、分母或两者都含有代数表达式的分式，例如 (x+2)/(x-3)。在 IGCSE Edexcel 数学中，掌握代数分式对于化简表达式、解方程以及学习函数和微积分等更高级主题至关重要。本文将从基本化简到解复杂分式方程，提供清晰且逐步的指导。

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1. Understanding Algebraic Fractions | 理解代数分式

An algebraic fraction is a fraction with variables in the numerator, denominator, or both. They follow the same rules as numerical fractions but require algebraic manipulation such as factorising and expanding.

代数分式是分子或分母中含有变量的分式。它们遵循与数值分式相同的运算规则，但需要进行因式分解和展开等代数处理。

Before performing any operation, always check if the fraction can be simplified by cancelling common factors. This is only possible when the numerator and denominator are fully factorised.

在进行任何运算之前，务必检查分式是否可以通过约去公因式进行化简。只有在分子和分母完全分解因式后，才能进行约分。

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2. Factorising – The Essential First Step | 因式分解——必不可少的第一步

Most operations on algebraic fractions begin with factorising both the numerator and the denominator. Factorising transforms expressions like x² - 4 into (x-2)(x+2), revealing common factors that can be cancelled.

大多数代数分式的运算都从分子和分母的因式分解开始。因式分解将 x² - 4 等表达式转化为 (x-2)(x+2)，从而揭示可以约分的公因式。

Common factorising methods include taking out a common factor, difference of two squares, and factorising quadratic trinomials. For example, 2x² + 6x can be written as 2x(x + 3).

常见的因式分解方法包括提取公因式、平方差公式以及二次三项式的因式分解。例如，2x² + 6x 可以写成 2x(x + 3)。

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

To multiply algebraic fractions, multiply the numerators together and multiply the denominators together. However, it is always more efficient to factorise and cancel any common factors before multiplying.

代数分式相乘时，将分子相乘作为新分子，分母相乘作为新分母。但更高效的做法是先分解因式，约去公因式，再进行乘法运算。

1) → factorise: [(x-1)(x+1)]/x × 2x/(x+1) → cancel (x+1) and x → result = 2(x-1) = 2x - 2.

例如：(x² - 1)/(x) × (2x)/(x + 1) → 因式分解：[(x-1)(x+1)]/x × 2x/(x+1) → 约去 (x+1) 和 x → 结果为 2(x-1) = 2x - 2。

Always remember to state any restrictions on x, such as x ≠ 0, x ≠ -1, to avoid division by zero.

切记要注明 x 的限制条件，如 x ≠ 0、x ≠ -1，以避免分母为零。

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

Dividing by an algebraic fraction is equivalent to multiplying by its reciprocal. Flip the second fraction and then proceed as in multiplication, factorising and cancelling first.

除以一个代数分式等于乘以它的倒数。将第二个分式上下颠倒，然后按乘法进行，先因式分解再约分。

+3)/(x+4) × (x-4)(x+4)/(x-3) → cancel (x-3) and (x+4) → result = (x+3)(x-4) = x² - x - 12.

例题：(x² - 9)/(x + 4) ÷ (x - 3)/(x² - 16) → 颠倒：(x² - 9)/(x + 4) × (x² - 16)/(x - 3) → 因式分解：(x-3)(x+3)/(x+4) × (x-4)(x+4)/(x-3) → 约去 (x-3) 和 (x+4) → 结果为 (x+3)(x-4) = x² - x - 12。

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5. Adding and Subtracting – Finding a Common Denominator | 加减法——寻找公分母

To add or subtract algebraic fractions, you must first express them with a common denominator. The common denominator should be the lowest common multiple (LCM) of the individual denominators.

要对代数分式进行加减运算，必须先将其化为同分母。公分母应是各个分母的最小公倍数 (LCM)。

write each fraction: 2(x-1)/[x(x-1)] + 3x/[x(x-1)] = (2x-2+3x)/[x(x-1)] = (5x-2)/[x(x-1)].

将每个分母完全因式分解，以确定最小公倍数。例如，合并 2/(x) + 3/(x-1)，LCM 为 x(x-1)。然后改写每个分式：2(x-1)/[x(x-1)] + 3x/[x(x-1)] = (2x-2+3x)/[x(x-1)] = (5x-2)/[x(x-1)]。

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6. Adding and Subtracting – Worked Examples | 加减法——详细例题

te: 4/[(x-1)(x+1)] - (x-1)/[(x-1)(x+1)] = (4 - x + 1)/[(x-1)(x+1)] = (5 - x)/[(x-1)(x+1)].

考虑减法：4/(x² - 1) - 1/(x + 1)。将 x² - 1 分解为 (x-1)(x+1)，LCM 为 (x-1)(x+1)。改写：4/[(x-1)(x+1)] - (x-1)/[(x-1)(x+1)] = (4 - x + 1)/[(x-1)(x+1)] = (5 - x)/[(x-1)(x+1)]。

: a/b + c/d = (ad+bc)/bd. But factorisation first is safer to avoid unnecessary expansion.

当分母为数字或简单表达式时，可以使用十字相乘法：a/b + c/d = (ad+bc)/bd。但先进行因式分解更安全，可避免不必要的展开。

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7. Solving Equations with Algebraic Fractions | 解含代数分式的方程

equire clearing denominators. Multiply every term by the LCM of all denominators, which is (x-2)(x+2), to eliminate fractions.

如 1/(x-2) + 1/(x+2) = 1 这样的方程，需要去分母。将每一项乘以所有分母的 LCM，即 (x-2)(x+2)，以消去分式。

Step: (x+2) + (x-2) = (x-2)(x+2) → 2x = x² - 4 → x² - 2x - 4 = 0 → x = 1 ± √5

步骤：(x+2) + (x-2) = (x-2)(x+2) → 2x = x² - 4 → 移项：x² - 2x - 4 = 0 → 解二次方程：x = [2 ± √4+16]/2 = 1 ± √5。

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8. Extraneous Solutions and Domain Restrictions | 增根与定义域限制

After solving, always check that solutions do not make any original denominator zero. If a solution results in division by zero, it must be rejected as extraneous.

求出解后，务必检验解是否会使原方程中的分母为零。若导致分母为零，则必须舍去，因为它是增根。

In the example above, both 1+√5 and 1-√5 do not equal ±2, so they are valid. Always state domain restrictions early: x ≠ 2, x ≠ -2.

上例中，1+√5 和 1-√5 均不等于 ±2，因此有效。应尽早标注定义域限制：x ≠ 2, x ≠ -2。

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9. Harder Examples with Quadratic Denominators | 含二次分母的复杂例题

Solve: x/(x-3) + 2/(x² - 9) = 1/(x+3). First factorise x

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