📚 Mixed Exercise 1 | 综合练习1
Welcome to this comprehensive revision article for Mixed Exercise 1 from the Edexcel A Level Pure Mathematics Year 1/AS textbook. This exercise consolidates foundational algebraic techniques, including index laws, expanding brackets, factorising, rationalising denominators and simplifying algebraic fractions. Strengthening these core skills is essential as they underpin almost every topic in the A Level mathematics course. By working through the examples and explanations below, you will build confidence in tackling the mixed problem sets and be well prepared for assessment-style questions.
欢迎来到这篇针对Edexcel A Level纯数学第一册/AS教材综合练习1的全面复习文章。本练习整合了基础的代数技巧，包括指数定律、去括号、因式分解、分母有理化以及代数分式的化简。巩固这些核心技能至关重要，因为它们几乎支撑着A Level数学课程中的每一个主题。通过学习下面的示例与讲解，你将逐步建立起解决综合题型所需的信心，并为应对考试风格的题目做好充分准备。
1. Laws of Indices | 指数定律
h as (2a³b)⁴ or evaluate numbers raised to fractional and negative powers. Remember the key rules: aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁰ = 1, a⁻ⁿ = 1/aⁿ and a¹/ⁿ = √a.
指数定律使我们能够高效地处理含有幂的表达式。在综合练习1中，你需要化简如 (2a³b)⁴ 这样的表达式，或计算带有分数指数与负指数的数值。请牢记关键法则：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，(aᵐ)ⁿ = aᵐⁿ，a⁰ = 1，a⁻ⁿ = 1/aⁿ 以及 a¹/ⁿ = √a。
result is 16a¹²b⁴. Always be careful with coefficients and apply the power to every part of the product.
例如，化简 (2a³b)⁴ 需要你将括号内的每个因式分别进行四次方运算：2⁴ = 16，(a³)⁴ = a¹²，b⁴ 保持不变。最终结果为 16a¹²b⁴。处理系数时务必仔细，并将幂运算应用到乘积的每一个部分。
(2a³b)⁴ = 16a¹²b⁴
When dealing with fractional or negative indices, break the problem into steps. For 8⁻²/³, first write it as 1/8²/³, then evaluate the cube root of 8 (which is 2) and square it to get 4, so the answer is 1/4. In Mixed Exercise 1, combining these rules accurately is vital for success.
在处理分数或负指数时，应分步进行。对于 8⁻²/³，首先改写为 1/8²/³，然后求8的立方根（得2），再将其平方得4，因此答案为1/4。在综合练习1中，准确组合运用这些法则是成功的关键。
2. Expanding Brackets | 去括号
Expanding brackets involves multiplying each term inside a bracket by the term outside, then simplifying the result. Mixed Exercise 1 tests your ability to expand expressions such as 3x(2x² – 5x + 1) and to handle products of two binomials like (2x + 3)(x – 4).
去括号是指将括号内的每一项与括号外的项相乘，然后化简结果。综合练习1考查你展开如 3x(2x² – 5x + 1) 这样的表达式，以及处理两个二项式乘积如 (2x + 3)(x – 4) 的能力。
or a single bracket, multiply the term outside by each term inside: 3x × 2x² = 6x³, 3x × (–5x) = –15x², and 3x × 1 = 3x. The expanded form is 6x³ – ys check your signs, as a common mistake is to lose a negative sign during multiplication.
对于单个括号，将括号外的项分别乘以括号内的每一项：3x × 2x² = 6x³，3x × (–5x) = –15x²，以及 3x × 1 = 3x。展开后的形式是 6x³ – 15x² + 3x。务必检查符号，因为在乘法中漏掉负号是一个常见错误。
n expanding two binomials, use the FOIL method (First, Outer, Inner, Last). For (2x + 3)(x – 4): First = 2x × x = 2x², Outer = 2x × (–4) = –8x, Inner = 3 × x = 3x, Last = 3 × (–4) = –12. Combine like terms: –8x + 3x = –5x, giving 2x² – 5x – counter expansions involving three brackets, but the structured approach remains the same.
当展开两个二项式时，使用FOIL方法（首项、外项、内项、末项）。对 (2x + 3)(x – 4)：首项 = 2x × x = 2x²，外项 = 2x × (–4) = –8x，内项 = 3 × x = 3x，末项 = 3 × (–4) = –12。合并同类项：–8x + 3x = –5x，最终得到 2x² – 5x – 12。在综合练习1中你还会遇到涉及三个括号的展开，但结构化的方法依然不变。
3. Factorising Simple Expressions | 简单表达式的因式分解
Factorising is the reverse of expanding; you look for a common factor in all terms and take it outside a bracket. In Mixed Exercise 1, you might need to factorise expressions such as 14x³y² – 21x²y³. The highest common factor here is 7x²y², so the factorised form is 7x²y²(2x – 3y).
因式分解是展开的逆运算；你需要找出所有项的公因子，并将其提到括号外。在综合练习1中，你可能需要分解如 14x³y² – 21x²y³ 的表达式。这里的最大公因子是 7x²y²，因此因式分解形式为 7x²y²(2x – 3y)。
Always check your factorisation by expanding back. Additionally, be prepared to factorise expressions where the common factor is a single variable: for 5x⁴ – 10x² + 15x, the factor 5x can be taken out, leaving 5x(x³ – 2x + 3). This skill is a prerequisite for simplifying algebraic fractions and solving equations later.
务必通过重新展开来检验你的因式分解结果。此外，还要能分解公因子为单个变量的表达式：对于 5x⁴ – 10x² + 15x，可提取公因子 5x，得到 5x(x³ – 2x + 3)。这项技能是后续化简代数分式和解方程的前提。
4. Factorising Quadratics | 二次三项式的因式分解
cs where a = 1, look for two numbers that multiply to c and add to b. For example, x² + 7x + 12 factorises to (x + 3)(x + 4) because 3 × 4 = 12 and 3 + 4 = 7.
二次三项式的因式分解是综合练习1中考查最多的技能之一。对于形如 x² + bx + c 或 ax² + bx + c 的二次式，当 a = 1 时，需找到两个数字，使它们的乘积为 c，和为 b。例如，x² + 7x + 12 可分解为 (x + 3)(x + 4)，因为 3 × 4 = 12 且 3 + 4 = 7。
rite 7x as 6x + x, then group: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Mixed Exercise 1 will challenge you to recognise the appropriate method quickly.
当 a 不等于1时，可以使用分组分解法。考虑 2x² + 7x + 3。先将 a 与 c 相乘：2 × 3 = 6。找出乘积为6、和为7的两个数：6 和 1。将 7x 改写为 6x + x，然后分组：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。综合练习1将考验你迅速识别并选取恰当方法的能力。
5. Rationalising Denominators | 分母有理化
Rationalising the denominator means eliminating surds from the bottom of a fraction. In Mixed Exercise 1, you will meet two types: denominators containing a single surd like √5, and those containing a sum or difference such as 2 + √3. For a single surd denominator, multiply numerator and denominator by that surd: for 3/√5, multiply by √5/√5 to get (3√5)/5.
分母有理化是指消去分数分母中的根式。在综合练习1中，你会遇到两种类型：分母只含有一个根式，如 √5；以及分母含有和或差，如 2 + √3。对于单根式分母，将分子和分母同乘该根式：对于 3/√5，同乘 √5/√5 得到 (3√5)/5。
For denominators like 5/(3 – √2), multiply numerator and denominator by the conjugate (3 + √2). The denominator becomes (3)² – (√2)² = 9 – 2 = 7, a rational number. The expression simplifies to (5(3 + √2))/7 = (15 + 5√2)/7. This skill is essential as rationalised forms are the standard way to present final answers.
对于 5/(3 – √2) 这类分母，将分子分母同乘共轭式 (3 + √2)。分母变为 (3)² – (√2)² = 9 – 2 = 7，成为一个有理数。该式化简为 (5(3 + √2))/7 = (15 + 5√2)/7。这项技能必不可少，因为有理化后的形式是呈现最终答案的标准方式。
6. Algebraic Fractions: Simplifying | 代数分式的化简
Simplifying algebraic fractions often requires factorising the numerator and denominator first, then cancelling common factors. Mixed Exercise 1 includes expressions like (x² – 9)/(x² – 2x – 3). The numerator is a difference of squares: (x + 3)(x – 3). The denominator factorises to (x – 3)(x + 1). Cancel the (x – 3) factor to obtain (x + 3)/(x + 1), provided x ≠ 3.
化简代数分式通常需要先对分子和分母进行因式分解，然后约去公因子。综合练习1包含如 (x² – 9)/(x² – 2x – 3) 的表达式。分子是一个平方差：(x + 3)(x – 3)。分母可分解为 (x – 3)(x + 1)。约去 (x – 3) 因子，得到 (x + 3)/(x + 1)，其中 x ≠ 3。
Always state any restrictions on x to ensure the original fraction remains defined. In this example, x cannot be 3 or –1. Neglecting to identify these restrictions can lead to loss of marks, as the question may explicitly ask for the simplified fraction together with its domain.
务必注明对 x 的限制条件，以确保原分式有定义。在本例中，x 不能为 3 或 –1。忽略注明这些限制条件可能会导致失分，因为题目可能明确要求给出化简后的分式及其定义域。
7. Algebraic Fractions: Adding and Subtracting | 代数分式的加减运算
Adding and subtracting algebraic fractions involve finding a common denominator, just as with numerical fractions. In Mixed Exercise 1, you may need to simplify expressions such as 2/(x + 1) + 3/(x – 2). The common denominator is (x + 1)(x – 2). Rewrite as (2(x – 2) + 3(x + 1)) / ((x + 1)(x – 2)).
代数分式的加减运算需要先找到公分母，这与数值分数的处理方式相同。在综合练习1中，你可能需要化简诸如 2/(x + 1) + 3/(x – 2) 的表达式。公分母是 (x + 1)(x – 2)。将式子改写为 (2(x – 2) + 3(x + 1)) / ((x + 1)(x – 2))。
Expand the numerator: 2x – 4 + 3x + 3 = 5x – 1. The result is (5x – 1)/((x + 1)(x – st apply to every term in the second numerator when combining over the common denominator.
将分子展开：2x – 4 + 3x + 3 = 5x – 1。结果为 (5x – 1)/((x + 1)(x – 2))。通常允许将分母保留为因式分解形式。在进行减法运算时，要格外注意负号：在通分合并时，减号必须作用于第二个分子的每一项。
8. Mixed Problem Solving | 综合应用与解题
Mixed Exercise 1 brings all the above skills together in multi-step problems. A typical question might ask: "Simplify fully (2x² – 8)/(x² + x – 2) ÷ (x – 2)/(x² + 3x + 2)." You must first factorise each algebraic fraction: 2(x – 2)(x + 2)/((x + 2)(x – 1)) and (x – 2)/((x + 2)(x + 1)).
综合练习1将上述所有技巧整合到多步骤问题中。一道典型题目可能会要求：“完全化简 (2x² – 8)/(x² + x – 2) ÷ (x – 2)/(x² + 3x + 2)。”你必须先对每个代数分式进行因式分解：2(x – 2)(x + 2)/((x + 2)(x – 1)) 以及 (x – 2)/((x + 2)(x + 1))。
a fraction is equivalent to multiplying by its reciprocal. So the expression becomes [2(x – 2)(x + 2)/((x + 2)(x – 1))] × [((x + 2)(x + 1))/(x – 2)]. Cancel common factors systematically. The (x + 2) cancels, the (x – 2) cancels, leaving 2(x + 1)/(x – 1), with x ≠ –2, 2, 1, –1. Structured cancellation prevents errors.
记住，除以一个分式等于乘以其倒数。因此表达式变为 [2(x – 2)(x + 2)/((x + 2)(x – 1))] × [((x + 2)(x + 1))/(x – 2)]。系统地约去公因子。(x + 2) 可约去，(x – 2) 可约去，最终得到 2(x + 1)/(x – 1)，且 x ≠ –2, 2, 1, –1。结构化的约分有助于避免错误。
9. Common Mistakes and How to Avoid Them | 常见错误与避错方法
stead of factors. For instance, in (x² + 4x)/(x²), a common mistake is to cancel the x² to get 4x. Instead, factorise the numerator: x(x + 4)/x² = (x + 4)/x, only after cancelling a common factor of x.
即使是自信的学生，也常因符号错误、忘记对系数进行幂运算或错误地消去项（而非因子）而在综合练习1中失分。例如，在 (x² + 4x)/(x²) 中，一个常见错误是直接将 x² 约去得到 4x。正确的做法是先将分子因式分解：x(x + 4)/x²，约去公因子 x 后得到 (x + 4)/x。
Another trap is mishandling the difference of squares. Always recognise that x² – 9 is (x – 3)(x + 3) and not (x – 3)². Take time to write each step clearly, and double-check by substituting a small value for x (other than excluded values) into both original and simplified expressions to verify they are equivalent.
另一个陷阱是错误地处理平方差。务必识别出 x² – 9 应分解为 (x – 3)(x + 3)，而不是 (x – 3)²。请花一些时间清晰地写出每一步，并通过代入一个较小的 x 值（排除被限制的值）到原式和化简后的式中进行双重检验，确认两者等价。
10. Exam-Style Tips for Mixed Exercise 1 | 综合练习1的考试技巧
In the actual Edexcel exam, questions similar to Mixed Exercise 1 often appear as short, standalone items in Paper 1 or as part of longer modelling problems. Always present your final answer in its simplest form, with the denominator rationalised and all algebraic fractions reduced. If the question asks for the value of a constant, make sure you substitute back to check.
在实际的Edexcel考试中，类似综合练习1的题目往往以简短的独立题形式出现在试卷1中，或作为较长的建模题的一部分。务必以最简形式给出最终答案，分母要完成有理化，所有代数分式都应约至最简。如果题目要求求出某个常数的值，请务必代回原式进行检验。
e highest common factor. If you get stuck, break the problem into smaller steps and write out what you do know. Marks are awarded for correct method even if the final answer contains a slip.
时间管理至关重要。在混合题型中，要熟悉因式分解的捷径，尤其是当二次项系数 a=1 时，并高效地识别出最大公因子。如果卡住了，可将问题分解为更小的步骤，并写下你所知道的部分。即便最终答案出现小错，正确的方法仍能获得分数。
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