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Cambridge IGCSE Mathematics: Algebra and Functions Core Revision | 剑桥 IGCSE 数学:代数与函数核心复习

2026-08-21 13:39 · 本文部分内容由AI辅助创作

📚 Cambridge IGCSE Mathematics: Algebra and Functions Core Revision | 剑桥 IGCSE 数学:代数与函数核心复习

This revision guide covers essential algebra and function topics from the Cambridge IGCSE Mathematics coursebook (ISBN 9781108760027), focusing on the first half of the syllabus. It is designed for both Core and Extended candidates preparing for assessment.

本复习指南涵盖剑桥 IGCSE 数学教材(ISBN 9781108760027)前半部分中代数与函数的核心主题,适合核心课程和扩展课程考生备考使用。

1. Simplifying Algebraic Expressions | 化简代数表达式

² - a² = a². Always identify terms with exactly the same variable part before simplifying.

合并同类项,即把系数相加或相减。例如,3x + 5x = 8x,2a² - a² = a²。化简前必须准确识别具有完全相同变量部分的项。

dices: xᵐ × xⁿ = xᵐ⁺ⁿ. When raising a power to a power, multiply the indices: (xᵐ)ⁿ = xᵐⁿ.

使用分配律去括号:a(b + c) = ab + ac。例如,4(2x - 3) = 8x - 12。同底数幂相乘时,指数相加:xᵐ × xⁿ = xᵐ⁺ⁿ。幂的乘方时,指数相乘:(xᵐ)ⁿ = xᵐⁿ。


2. Solving Linear Equations | 解一次方程

obtain x = 4. Always check your answer by substituting it back into the original equation.

运用逆运算把未知量单独留在等号一边。例如:解方程 3x + 7 = 19。先两边同时减去 7,得到 3x = 12,再两边同时除以 3,得到 x = 4。务必把答案代回原方程检验。

x = 16, so x = 8. If the unknown appears on both sides, collect like terms before solving.

对于含括号的方程,先去括号。例如:2(x - 3) = 10 化为 2x - 6 = 10,再得 2x = 16,所以 x = 8。如果未知量出现在等号两边,先移项合并同类项再求解。


3. Solving Quadratic Equations | 解二次方程

= 0, so x = 2 or x = 3. The product must equal zero for at least one factor to equal zero.

因式分解是首选方法。例如:x² - 5x + 6 = 0 可写成 (x - 2)(x - 3) = 0,所以 x = 2 或 x = 3。只有当乘积为零时,至少一个因式为零。

not possible, use the quadratic formula. For ax² + bx + c = 0, the solutions are given by:

x = (−b ± √b² − 4ac) ÷ (2a)

当无法因式分解时,使用求根公式。对于 ax² + bx + c = 0,解为:

x = (−b ± √b² − 4ac) ÷ (2a)

The discriminant b² - 4ac tells you the nature of the roots: positive gives two distinct real roots, zero gives one repeated root, and negative gives no real roots.

判别式 b² - 4ac 可以判断根的性质:大于零有两个不相等的实根,等于零有一个重根,小于零没有实根。


4. Simultaneous Equations | 联立方程组

le: 2x + y = 7 and x - y = 2. Adding gives 3x = 9, so x = 3, then substitute to get y = 1.

消元法通过相加或相减消去一个变量。例如:2x + y = 7 与 x - y = 2。两式相加得 3x = 9,所以 x = 3,再代入得 y = 1。

2x + 1) = 10, so 3x = 9, x = 3, and y = 7. Always write the solution as x = ..., y = ... .

代入法用一个变量的表达式替换另一个变量。例如:y = 2x + 1 与 x + y = 10。代入得 x + (2x + 1) = 10,所以 3x = 9,x = 3,y = 7。答案必须写成 x = ..., y = ... 的形式。


5. Inequalities | 不等式

Solve inequalities like equations, but reverse the inequality sign when multiplying or dividing by a negative number. Example: -2x < 8 becomes x > -4 after dividing both sides by -2.

解不等式与解方程类似,但当两边同乘或同除以一个负数时,必须反转不等号方向。例如:-2x < 8 两边同除以 -2 后变为 x > -4。

ed circle for inclusive inequalities like ≤ or ≥. Always show the direction with an arrow.

用数轴表示解集。严格不等式(如 < 或 >)用空心圆圈,包含不等号的不等式(如 ≤ 或 ≥)用实心圆圈。必须用箭头标明方向。


6. Sequences | 数列

is the common difference. Example: 3, 7, 11, 15... has nth term 3 + (n - 1) × 4 = 4n - 1.

等差数列相邻两项的差为常数。第 n 项公式为 a + (n - 1)d,其中 a 是首项,d 是公差。例如:3, 7, 11, 15... 的第 n 项为 3 + (n - 1) × 4 = 4n - 1。

For a quadratic sequence, the second differences are constant. The nth term has the form an² + bn + c. Find a from half the second difference, then use known terms to find b and c.

对于二次数列,二阶差为常数。第 n 项形式为 an² + bn + c。用二阶差的一半确定 a,再代入已知项求 b 和 c。


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7. Functions and Function Notation | 函数与函数记号

function maps each input to exactly one output. In function notation, f(x) = 2x + 3 means 'f of x equals 2x + 3'. To evaluate, substitute the input: f(4) = 2(4) + 3 = 11.

函数将每个输入唯一地映射到一个输出。在函数记号中,f(x) = 2x + 3 表示“f 作用于 x 等于 2x + 3”。求值时代入输入:f(4) = 2(4) + 3 = 11。

n fg(x) means apply g first, then f. If f(x) = x² and g(x) = x + 1, then fg(x) = (x + 1)².

反函数逆转映射关系。对于 f(x) = 2x + 3,设 y = 2x + 3,交换 x 与 y 后解出 y:f⁻¹(x) = (x - 3) ÷ 2。复合函数 fg(x) 表示先作用 g,再作用 f。若 f(x) = x² 且 g(x) = x + 1,则 fg(x) = (x + 1)²。


8. Graphs of Functions | 函数图像

the y-intercept. For example, y = 2x - 1 has gradient 2 and crosses the y-axis at (0, -1).

一次函数的形式为 y = mx + c,其中 m 是斜率,c 是 y 轴截距。例如,y = 2x - 1 的斜率为 2,与 y 轴交于 (0, -1)。

a maximum point. The vertex can be found by completing the square or using x = -b ÷ (2a).

二次函数 y = ax² + bx + c 的图像是一条抛物线。若 a > 0,抛物线开口向上,有最低点;若 a < 0,开口向下,有最高点。顶点可通过配方法或 x = -b ÷ (2a) 求得。


9. Direct and Inverse Proportion | 正比例与反比例

Example: if y is directly proportional to x and y = 10 when x = 2, then k = 5 and y = 5x.

两个量成正比例是指它们的比值为常数:y = kx,所以 y ÷ x = k。图像是过原点的一条直线。例如:若 y 与 x 成正比例,且 x = 2 时 y = 10,则 k = 5,函数为 y = 5x。

inversely proportional to x and y = 6 when x = 2, then k = 12. When x = 4, y = 12 ÷ 4 = 3.

两个量成反比例是指它们的乘积为常数:y = k ÷ x,所以 xy = k。图像是双曲线。例如:若 y 与 x 成反比例,且 x = 2 时 y = 6,则 k = 12。当 x = 4 时,y = 12 ÷ 4 = 3。


10. Algebraic Fractions and Indices | 代数分式与指数

n factors. Example: (x² - 9) ÷ (x - 3) = (x + 3)(x - 3) ÷ (x - 3) = x + 3, provided x ≠ 3.

化简代数分式时,先把分子和分母因式分解,再约去公因式。例如:(x² - 9) ÷ (x - 3) = (x + 3)(x - 3) ÷ (x - 3) = x + 3,其中 x ≠ 3。

Negative indices mean reciprocal: x⁻¹ = 1 ÷ x. Fractional indices represent roots: x1/2 = √x and x2/3 = ∛ ply the rules xᵐ × xⁿ = xᵐ⁺ⁿ and xᵐ ÷ xⁿ = xᵐ⁻ⁿ when simplifying expressions with indices.

负指数表示倒数:x⁻¹ = 1 ÷ x。分数指数表示根式:x1/2 = √x,x2/3 = ∛(x²)。化简含指数的表达式时,始终运用 xᵐ × xⁿ = xᵐ⁺ⁿ 和 xᵐ ÷ xⁿ = xᵐ⁻ⁿ。


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