📚 Review Exercise 1 for Edexcel A-Level Maths | Edexcel A-Level数学复习练习1
Review Exercise 1 in Edexcel A-Level Mathematics usually checks the key skills from the first half of Pure Mathematics Year 1: algebraic manipulation, quadratics, equations and inequalities, straight lines, circles and graph transformations. This article revises the main ideas and common question types so you can work through the exercise confidently.
复习练习1通常考查纯数学第一册前半部分的核心技能：代数运算、二次函数、方程与不等式、直线、圆以及图像变换。本文整理主要知识点和常见题型，帮助你更自信地完成练习。
1. Algebraic Expressions and Index Laws | 代数表达式与指数律
Simplifying expressions correctly is the foundation of most exam questions. You need to be confident with the index laws: aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, and (aᵐ)ⁿ = aᵐⁿ. Also recall that a⁰ = 1 and a⁻ⁿ = 1/aⁿ. These rules often appear when you simplify algebraic fractions or solve equations involving powers.
正确化简代数式是大多数考题的基础。你需要熟练运用指数律：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，以及 (aᵐ)ⁿ = aᵐⁿ。还要记住 a⁰ = 1，a⁻ⁿ = 1/aⁿ。这些法则经常出现在化简代数分式或解含幂的方程中。
Surds can be simplified by writing the number under the root as a product of a square number and another factor, for example √50 = √25 × √2 = 5√2. Rationalising the denominator is tested frequently, so remember that 1/√a can be multiplied by √a/√a to give √a/a.
根式可把被开方数写成平方数与另一因数的乘积来化简，例如 √50 = √25 × √2 = 5√2。分母有理化也经常出现，所以要记住 1/√a 可以乘以 √a/√a，得到 √a/a。
aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ
2. Expanding and Factorising Quadratics | 二次式展开与因式分解
Expand two brackets using the distributive law: (x + a)(x + b) = x² + (a + b)x + ab. Watch negative signs when subtracting brackets, because mistakes often come from losing a minus sign. For example, (x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15.
用乘法分配律展开两个括号：(x + a)(x + b) = x² + (a + b)x + ab。减去括号时要特别注意负号，因为错误往往来自漏掉负号。例如 (x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15。
Factorising is the reverse process: find two numbers that multiply to the constant term and add to the coefficient of x. For example, x² − 7x + 10 = (x − 2)(x − 5). The difference of two squares, a² − b² = (a − b)(a + b), is a very common pattern and should be recognised quickly.
因式分解是展开的逆过程：找出两个数，使它们相乘等于常数项、相加等于 x 的系数。例如 x² − 7x + 10 = (x − 2)(x − 5)。平方差 a² − b² = (a − b)(a + b) 是非常常见的结构，应当快速识别。
- x² + 5x + 6 = (x + 2)(x + 3)
- x² − 9 = (x − 3)(x + 3)
- 2x² + 7x + 3 = (2x + 1)(x + 3)
3. Completing the Square | 配方法
Completing the square rewrites a quadratic as a(x + p)² + q or a(x − p)² + q. This form gives the turning point (−p, q) or (p, q), and helps solve equations and sketch graphs. For x² + 6x + 2, halve the x coefficient: (x + 3)² − 9 + 2 = (x + 3)² − 7.
配方法把二次式写成 a(x + p)² + q 或 a(x − p)² + q 的形式。这个形式可以直接读出顶点 (−p, q) 或 (p, q)，也有助于解方程和画图像。例如 x² + 6x + 2，先取 x 系数的一半：(x + 3)² − 9 + 2 = (x + 3)² − 7。
If the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 5, write it as 2(x² + 4x) + 5, then complete the square inside the bracket: 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3. The turning point is (−2, −3).
若 x² 的系数不是 1，需先把它提出来。例如 2x² + 8x + 5，写成 2(x² + 4x) + 5，再在括号内配方：2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。顶点为 (−2, −3)。
x² + bx + c = (x + b/2)² − (b/2)² + c
4. Quadratic Formula and the Discriminant | 二次公式与判别式
The quadratic formula solves ax² + bx + c = 0. It is given in the Edexcel formula booklet, but you should be able to use it quickly and accurately.
二次公式可以解 ax² + bx + c = 0。它在 Edexcel 公式表中给出，但你应该能够快速准确地使用它。
x = (−b ± √(b² − 4ac)) ÷ (2a)
The discriminant D = b² − 4ac tells you the nature of the roots. D > 0 gives two distinct real roots, D = 0 gives one repeated real root, and D < 0 gives no real roots. Edexcel often asks you to find the set of values of k for which a quadratic has real roots, so set D ≥ 0 and solve the resulting inequality.
判别式 D = b² − 4ac 告诉你根的性质。D > 0 有两个不相等的实根，D = 0 有一个重根，D < 0 没有实根。Edexcel 经常要求找使二次方程有实根的 k 的取值范围，因此要令 D ≥ 0，再解所得不等式。
For example, x² + kx + 9 = 0 has real roots when k² − 36 ≥ 0, so k ≤ −6 or k ≥ 6. Always check whether the question says “real roots”, “distinct real roots” or “no real roots”, because this changes the inequality sign.
例如，x² + kx + 9 = 0 有实根时需要 k² − 36 ≥ 0，因此 k ≤ −6 或 k ≥ 6。一定要看清题目要求是“有实根”、“有两个不相等的实根”还是“没有实根”，这会改变不等号的方向。
5. Linear and Quadratic Inequalities | 线性与二次不等式
Solve linear inequalities like equations, but reverse the inequality sign when multiplying or dividing by a negative number. For example, −3x < 12 becomes x > −4 after dividing by −3. Always show the sign change clearly to avoid losing marks.
线性不等式可按方程来解，但当乘以或除以负数时，要改变不等号的方向。例如 −3x < 12 除以 −3 后变为 x > −4。一定要清晰地展示变号过程，以免丢分。
For quadratic inequalities, first rearrange to make the right side 0, factorise the left side, find the critical values, then use a sign diagram or graph to choose intervals. Strict inequalities use open circles; inclusive inequalities use closed circles. A quick sketch of y = ax² + bx + c can help you see which region is positive or negative.
二次不等式要先把右边化为 0，再因式分解左边，求出临界值，然后用符号表或图像选择区间。严格不等式用空心点，包含等号用实心点。快速画出 y = ax² + bx + c 的草图可以帮助你判断哪一段为正、哪一段为负。
x² − 3x − 10 < 0 → (x − 5)(x + 2) < 0 → −2 < x < 5
6. Simultaneous Equations | 联立方程
Two linear equations can be solved by elimination or substitution. If one equation is linear and one is quadratic, substitute the linear expression into the quadratic. This usually gives a quadratic in one variable, which may have two solutions, and each x value must be paired with its corresponding y value.
两个线性方程可用消元法或代入法求解。如果一个方程是线性、另一个是二次，就把线性表达式代入二次方程。这通常会化成一个变量的二次方程，可能有两个解，每个 x 值都要配出对应的 y 值。
Never stop at x alone. If you find x = 2 and x = −1, use the linear equation to find the matching y values. Also check both solutions in the original equations, because substitution can sometimes introduce extra solutions or cause arithmetic errors.
不要只解出 x 就停止。如果你求得 x = 2 和 x = −1，要用线性方程求出对应的 y 值。还要把两个解代回原方程检验，因为代入过程中有时会产生多余解或计算错误。
7. Straight Line Graphs | 直线图像
The equation y = mx + c has gradient m and y-intercept c. Two points (x₁, y₁) and (x₂, y₂) give gradient m = (y₂ − y₁) ÷ (x₂ − x₁). Parallel lines have equal gradients; perpendicular lines have gradients whose product is −1.
y = mx + c 的斜率为 m，y 轴截距为 c。两点 (x₁, y₁) 和 (x₂, y₂) 的斜率 m = (y₂ − y₁) ÷ (x₂ − x₁)。平行线斜率相等；垂直线斜率的乘积为 −1。
You often need to write the equation of a line through a point with a given gradient using y − y₁ = m(x − x₁). If the gradient is not given directly, calculate it from two points or from a perpendicular line. Be ready to convert between y = mx + c and ax + by + c = 0.
你可能需要根据已知斜率和一点写出直线方程，使用 y − y₁ = m(x − x₁)。如果没有直接给出斜率，就从两点或垂直线求出斜率。还要能够在 y = mx + c 与 ax + by + c = 0 之间相互转换。
m = (y₂ − y₁) ÷ (x₂ − x₁), y − y₁ = m(x − x₁)
8. Circles | 圆
The equation of a circle with centre (a, b) and radius r is (x − a)² + (y − b)² = r². To find the centre and radius from expanded form, complete the square for both x and y. For example, x² + y² − 4x + 6y − 3 = 0 can be rewritten as (x − 2)² + (y + 3)² = 16, so the centre is (2, −3) and radius is 4.
圆心为 (a, b)、半径为 r 的圆的方程是 (x − a)² + (y − b)² = r²。要从一般式找出圆心和半径，需要对 x 和 y 分别配方。例如 x² + y² − 4x + 6y − 3 = 0 可改写为 (x − 2)² + (y + 3)² = 16，所以圆心为 (2, −3)，半径为 4。
A tangent is perpendicular to the radius at the point of contact, so tangent questions often use the negative reciprocal gradient. Find the gradient of the radius from the centre to the contact point, then take its negative reciprocal to find the tangent gradient, and finally use the point-gradient formula.
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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