📚 AQA Mathematics: Calculation Skills Bootcamp | AQA 数学：计算题专项训练
In AQA Mathematics, strong calculation skills are the foundation for success across pure and applied topics. This article provides a focused drilling session covering essential computational techniques from algebraic manipulation to complex numbers. Master these to boost your speed and accuracy in exams.
在 AQA 数学中，扎实的计算技能是纯数学与应用数学成功的基础。本文提供一次集中的专项训练，涵盖从代数化简到复数的关键计算技巧。掌握这些内容可以有效提升你的解题速度和准确率。
1. Simplifying Algebraic Expressions | 代数表达式化简
Collect like terms and use the distributive law carefully. Signs often cause errors, so expand brackets step by step. For instance, simplify 3(x+2) − 2(2x−1) by writing each product explicitly.
仔细合并同类项并运用分配律。符号常常导致错误，因此要逐步展开括号。例如，化简 3(x+2) − 2(2x−1)，先把每一项写出。
3(x+2) − 2(2x−1) = 3x + 6 − 4x + 2 = −x + 8
3(x+2) − 2(2x−1) = 3x + 6 − 4x + 2 = −x + 8
Expand two binomials by multiplying each term of the first bracket by every term of the second. Example: (2x−3)(x+4) yields 2x·x, 2x·4, −3·x and −3·4. Then combine the x-terms.
展开两个二项式时，将第一个括号中的每一项乘以第二个括号中的每一项。例如：(2x−3)(x+4) 得到 2x·x、2x·4、−3·x 和 −3·4，然后合并 x 项。
(2x−3)(x+4) = 2x² + 8x − 3x − 12 = 2x² + 5x − 12
(2x−3)(x+4) = 2x² + 8x − 3x − 12 = 2x² + 5x − 12
2. Solving Linear and Quadratic Equations | 解一次与二次方程
For linear equations, isolate the variable by performing the same operation on both sides. Always check your solution in the original equation. Solve 2x − 7 = 3x + 5 by moving x-terms to one side.
对于一次方程，通过对等号两边进行相同运算来分离变量。始终将解代入原方程检验。解 2x − 7 = 3x + 5 时，把含 x 的项移到一侧。
2x − 7 = 3x + 5 → −7 − 5 = 3x − 2x → −12 = x → x = −12
2x − 7 = 3x + 5 → −7 − 5 = 3x − 2x → −12 = x → x = −12
Quadratic equations can often be factorised. Set the expression to zero, find two numbers that multiply to ac and add to b, then solve the brackets. For x² − 5x + 6 = 0, the factors are (x−2)(x−3).
二次方程往往可以因式分解。令表达式等于零，找出两个数，其乘积为 ac 且和为 b，然后分别令因式等于零。对于 x² − 5x + 6 = 0，因式为 (x−2)(x−3)。
When a quadratic cannot be factorised, apply the quadratic formula. Memorise x = [−b ± √(b² − 4ac)] / (2a). Discriminant b²−4ac determines the nature of roots.
当二次方程无法因式分解时，应用求根公式。牢记 x = [−b ± √(b² − 4ac)] / (2a)。判别式 b²−4ac 决定根的性质。
3. Manipulating Surds and Indices | 根式与指数运算
Simplify surds by extracting square factors. √12 = √(4×3) = 2√3. Always express surds in their simplest form. To rationalise a denominator like 1/(√2+1), multiply numerator and denominator by the conjugate √2−1.
通过提取平方因子化简根式。√12 = √(4×3) = 2√3。始终将根式表为最简形式。若要分母有理化，如 1/(√2+1)，将分子分母同乘共轭根式 √2−1。
1/(√2+1) = (√2−1)/[(√2+1)(√2−1)] = (√2−1)/(2−1) = √2−1
1/(√2+1) = (√2−1)/[(√2+1)(√2−1)] = (√2−1)/(2−1) = √2−1
Indices follow strict rules: aⁿ × aᵐ = aⁿ⁺ᵐ, (aⁿ)ᵐ = aⁿᵐ, and a⁻ⁿ = 1/aⁿ. For example, simplify (x³y⁻²)². Square each factor: (x³)² = x⁶, (y⁻²)² = y⁻⁴, so overall x⁶y⁻⁴ or x⁶/y⁴.
指数遵循严格的法则：aⁿ × aᵐ = aⁿ⁺ᵐ，(aⁿ)ᵐ = aⁿᵐ，以及 a⁻ⁿ = 1/aⁿ。例如化简 (x³y⁻²)²，将每个因子平方：(x³)² = x⁶，(y⁻²)² = y⁻⁴，结果为 x⁶y⁻⁴ 或 x⁶/y⁴。
4. Differential Calculus: Rules and Applications | 导数运算：法则与应用
The power rule states that d/dx (xⁿ) = nxⁿ⁻¹. Apply term by term. For a sum like f(x) = 4x³ − 2x + 7, differentiate each part to get f'(x) = 12x² − 2.
幂法则表明 d/dx (xⁿ) = nxⁿ⁻¹。逐项求导即可。对于如 f(x) = 4x³ − 2x + 7 的和式，分别求导得 f'(x) = 12x² − 2。
The chain rule handles composite functions: if y = [u(x)]ⁿ, then dy/dx = n·u(x)ⁿ⁻¹·u'(x). Example: y = (3x²+1)⁵. Let u = 3x²+1, then y’ = 5u⁴·6x = 30x(3x²+1)⁴.
链式法则处理复合函数：若 y = [u(x)]ⁿ，则 dy/dx = n·u(x)ⁿ⁻¹·u'(x)。例如 y = (3x²+1)⁵，令 u = 3x²+1，则 y’ = 5u⁴·6x = 30x(3x²+1)⁴。
Product rule: d/dx (uv) = u’v + uv’. Quotient rule: d/dx (u/v) = (u’v − uv’)/v². These are essential for products and fractions of functions.
乘积法则：d/dx (uv) = u’v + uv’。商法则：d/dx (u/v) = (u’v − uv’)/v²。这些对于函数乘积和分式求导至关重要。
5. Integral Calculus: Finding Antiderivatives | 积分运算：求原函数
Indefinite integration reverses differentiation: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1. Always add the constant of integration. For example, ∫ (3x² + 4) dx = x³ + 4x + C.
不定积分是求导的逆运算：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，其中 n ≠ −1。永远加上积分常数。例如 ∫ (3x² + 4) dx = x³ + 4x + C。
Definite integrals evaluate the area under a curve between limits a and b. Compute ∫₁² (2x+3) dx by integrating to x²+3x, then substituting x=2 and x=1: [4+6] − [1+3] = 10−4 = 6.
定积分计算曲线在区间 [a, b] 下的面积。计算 ∫₁² (2x+3) dx，先积分得 x²+3x，然后代入 x=2 和 x=1：[4+6] − [1+3] = 10−4 = 6。
Integrating basic functions: ∫ eˣ dx = eˣ + C; ∫ 1/x dx = ln|x| + C; ∫ cos x dx = sin x + C. Keep these standard results at your fingertips.
基本函数积分：∫ eˣ dx = eˣ + C；∫ 1/x dx = ln|x| + C；∫ cos x dx = sin x + C。熟记这些标准结果。
6. Trigonometric Identities and Equations | 三角恒等式与方程
Core identities: sin²θ + cos²θ = 1, tanθ = sinθ/cosθ. Use these to simplify expressions and solve equations. For instance, solve sin2θ = 1/2 for 0° ≤ θ ≤ 360°. First, let 2θ = X, so sin X = 1/2 gives X = 30°, 150°, 390°, 510°. Then θ = 15°, 75°, 195°, 255°.
核心恒等式：sin²θ + cos²θ = 1，tanθ = sinθ/cosθ。运用它们化简表达式和解方程。例如，在 0° ≤ θ ≤ 360° 内解 sin2θ = 1/2，令 2θ = X，则 sin X = 1/2 得 X = 30°、150°、390°、510°。于是 θ = 15°、75°、195°、255°。
Quadratic trig equations appear regularly. Solve 2cos²θ − cosθ − 1 = 0 as a quadratic in cosθ. Factor: (2cosθ+1)(cosθ−1)=0, so cosθ = −1/2 or cosθ = 1, giving θ = 120°, 240°, 0°, 360°.
二次三角方程经常出现。将 2cos²θ − cosθ − 1 = 0 视为 cosθ 的二次方程，因式分解得 (2cosθ+1)(cosθ−1)=0，故 cosθ = −1/2 或 cosθ = 1，解得 θ = 120°、240°、0°、360°。
7. Exponentials and Logarithms: Simplification and Equations | 指数与对数：化简与方程
Recall log properties: log a + log b = log(ab), log a − log b = log(a/b), and log aⁿ = n log a. The natural logarithm ln follows the same rules. Simplify ln 8 − ln 2 = ln(8/2) = ln 4.
回忆对数性质：log a + log b = log(ab)，log a − log b = log(a/b)，以及 log aⁿ
Published by TutorHao | Mathematics Revision Series | aleveler.com
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