📚 AS Mathematics: Inequalities Key Points | AS 数学：不等式 考点精讲
Inequalities are a fundamental topic in AS Mathematics, building on GCSE knowledge and introducing new techniques for solving quadratic, rational, and absolute value inequalities. This article highlights the key concepts, common pitfalls, and exam strategies needed to master inequalities at the AS level, ensuring you can confidently tackle any related question in your exams.
不等式是 AS 数学中的基础课题，它在 GCSE 知识的基础上进一步引入了求解二次不等式、分式不等式和绝对值不等式的新技巧。本文重点梳理关键概念、常见错误和应试策略，帮助你扎实掌握 AS 阶段的不等式，从容应对考试中的各类相关题目。
1. Inequality Symbols and Basic Properties | 不等式符号与基本性质
Inequalities use the symbols <, >, ≤, and ≥. The basic rules for manipulating inequalities are similar to equations, but with one crucial exception: multiplying or dividing both sides by a negative number reverses the inequality sign. Always keep this in mind when solving.
不等式使用符号 <、>、≤ 和 ≥。处理不等式的基本规则与方程类似，但有一个关键例外：当两边同时乘以或除以一个负数时，不等号方向必须反转。解题时务必时刻牢记这一点。
Adding or subtracting the same number from both sides does not change the inequality. Also, if you multiply or divide by a positive number, the direction remains the same. These properties allow us to isolate the variable.
两边同时加上或减去同一个数不会改变不等式的方向。同样，当乘以或除以正数时，方向保持不变。这些性质使我们能够将变量分离出来。
2. Solving Linear Inequalities | 求解线性不等式
Linear inequalities like 3x – 5 > 7 are solved by performing inverse operations exactly as in linear equations, remembering to flip the sign if multiplying or dividing by a negative. The solution set is usually expressed in set notation or interval notation, e.g., x > 4 or (4, ∞).
像 3x – 5 > 7 这样的线性不等式，可通过逆运算求解，方法与线性方程相同，但要注意乘以或除以负数时翻转不等号。解集通常用集合符号或区间符号表示，例如 x > 4 或 (4, ∞)。
Always check your final inequality by substituting a value from your solution set back into the original inequality. For example, if x > 4, try x = 5: 3(5) – 5 = 10, which is > 7, so it works.
始终将解集中的某个值代回原不等式进行检验。例如，若 x > 4，可尝试 x = 5：3(5) – 5 = 10，大于 7，符合要求。
3. Quadratic Inequalities and Critical Values | 二次不等式与临界值
To solve a quadratic inequality such as x² – 5x + 6 < 0, first find the critical values by solving the corresponding equation x² – 5x + 6 = 0. Factorising gives (x – 2)(x – 3) = 0, so the critical values are x = 2 and x = 3.
求解二次不等式如 x² – 5x + 6 < 0，首先要解对应的方程 x² – 5x + 6 = 0，找出临界值。因式分解得 (x – 2)(x – 3) = 0，因此临界值为 x = 2 和 x = 3。
These values divide the number line into three intervals: x < 2, 2 < x < 3, and x > 3. Test a point from each interval in the original inequality to find which intervals satisfy it. For x² – 5x + 6 < 0, the solution is 2 < x < 3, or (2, 3) in interval notation.
这些值将数轴分为三个区间：x < 2，2 < x < 3 和 x > 3。从每个区间中选取一个点代入原不等式进行验证，即可确定哪些区间满足条件。对于 x² – 5x + 6 < 0，解为 2 < x < 3，用区间符号表示为 (2, 3)。
Alternatively, sketch the graph of y = x² – 5x + 6, a U-shaped parabola intersecting the x-axis at 2 and 3. The inequality < 0 asks where the curve is below the x-axis, which occurs between the roots. This visual method is often quicker and less error-prone.
此外，也可以绘制 y = x² – 5x + 6 的图像，这是一条 U 形抛物线，与 x 轴交于 2 和 3 两点。不等式 < 0 要求找出曲线在 x 轴下方的部分，这部分正好位于两个根之间。这种图形方法通常更快捷且不易出错。
4. Using a Sign Table for Polynomial Inequalities | 用符号表解多项式不等式
For higher-degree polynomial inequalities like (x – 1)(x + 2)(x – 4) ≥ 0, a sign table (or number line method) is very effective. First, find all roots: x = –2, 1, and 4. List these in increasing order on a number line.
对于 (x – 1)(x + 2)(x – 4) ≥ 0 这样的高次多项式不等式，符号表（或数轴法）非常有效。首先找出所有根：x = –2、1 和 4。将它们按递增顺序排列在数轴上。
Determine the sign of the product in each interval by checking a test value. The sign generally alternates if all factors are linear and the product is expanded. The solution includes the intervals where the product is non-negative, and the endpoints where the expression equals zero (since it’s ≥).
通过代入检验值，确定各区间的乘积符号。如果所有因式均为线性且表达式已展开，符号通常会交替变化。解集包括乘积为非负数的区间，以及表达式等于零的端点（因为是不等式“≥”）。
5. Rational Inequalities and Avoiding Multiplication by Denominators | 分式不等式与避免分母直接相乘
Do not multiply both sides of a rational inequality by a denominator containing the variable without careful consideration of its sign. Instead, bring all terms to one side to form a single fraction compared to zero, e.g., (x+1)/(x–2) > 0.
在分式不等式中，不要轻易将含有变量的分母乘到另一侧，因为不清楚其正负。正确的做法是将所有项移到一侧，形成单分式与零比较，例如 (x+1)/(x–2) > 0。
Find the critical values from the numerator (x = –1) and denominator (x = 2). These split the number line into intervals. Test each interval to see where the fraction is positive. Remember x = 2 is a vertical asymptote and must be excluded from the solution set.
从分子 (x = –1) 和分母 (x = 2) 求出临界值。这些值将数轴划分为多个区间。逐一检验各区间内分式的符号。注意 x = 2 是垂直渐近线，必须从解集中剔除。
6. Absolute Value Inequalities and the Two-Case Method | 绝对值不等式与双情况法
Inequalities with absolute value, such as |2x – 3| < 5, are solved by splitting into two cases. The expression inside the absolute value must be between –5 and 5: –5 < 2x – 3 < 5. Solve this double inequality to get –1 < x < 4.
含有绝对值的不等式，例如 |2x – 3| < 5，可通过拆分为两种情况进行求解。绝对值内部的表达式必须介于 -5 和 5 之间：-5 < 2x – 3 < 5。解这个双不等式得到 -1 < x < 4。
For expressions like |x + 1| > 3, the quantity inside is either less than –3 or greater than 3. Therefore, x + 1 < –3 or x + 1 > 3, giving x < –4 or x > 2. Always express the final answer as a union of intervals.
对于 |x + 1| > 3 这类情况，绝对值内部的量要么小于 -3，要么大于 3。因此，x + 1 < –3 或 x + 1 > 3，解得 x < –4 或 x > 2。最终答案通常表示为区间的并集。
7. Simultaneous Linear Inequalities | 联合线性不等式
Solving simultaneous linear inequalities involves finding the region where both inequalities hold. Solve each inequality separately, then find the intersection of their solution sets. For example, solve 2x – 1 < 5 and 3x + 2 ≥ –1 to get x < 3 and x ≥ –1, so the combined solution is –1 ≤ x < 3.
求解联合线性不等式就是找出同时满足所有不等式的区域。分别求解每个不等式，然后找出它们解集的交集。例如，解 2x – 1 < 5 和 3x + 2 ≥ –1 得到 x < 3 且 x ≥ –1，因此综合解为 –1 ≤ x < 3。
Often these are presented as a double inequality, e.g., –1 ≤ 2x + 3 < 7. Solve by performing the same operation on all three parts: subtract 3, then divide by 2, giving –2 ≤ x < 2.
这类问题常常以双不等式形式出现，例如 –1 ≤ 2x + 3 < 7。此时可以对三部分同时进行相同操作来求解：先减 3，再除以 2，得到 –2 ≤ x < 2。
8. Graphical Representation of Inequalities | 不等式的图形表示
In the coordinate plane, linear inequalities like y > 2x + 1 are represented by shading a half-plane. Draw the boundary line y = 2x + 1 as a dashed line if the inequality is strict (> or <), or solid if it is ≤ or ≥. Then shade the appropriate side by testing a point, commonly (0,0).
在坐标平面上，y > 2x + 1 这样的线性不等式通过给半平面上色来表示。先画出边界线 y = 2x + 1，若不等式为严格不等号（> 或 <）则用虚线，若为 ≤ 或 ≥ 则用实线。然后通过代入一个点（通常用原点 (0,0)）检验，对正确的一侧进行上色。
Systems of inequalities, e.g., y < x + 2, x ≥ 0, y ≥ 1, require shading the intersection of all regions. The unshaded region or the region labelled R is often what the question asks you to identify and label clearly.
不等式组（例如 y < x + 2，x ≥ 0，y ≥ 1）需要给所有区域的交集上色。题目通常会要求标示并清楚地标记出未上色区域或标记为 R 的区域。
9. Set Notation and Interval Notation | 集合符号与区间符号
AS exams expect you to express solutions using both set notation {x : x > 3} and interval notation (3, ∞). Parentheses ( ) indicate the endpoint is not included, while square brackets [ ] indicate inclusion. For combined intervals, use the union symbol ∪, e.g., (−∞, 1) ∪ (4, ∞).
AS 考试要求能够使用集合符号 {x : x > 3} 和区间符号 (3, ∞) 来表示解。圆括号 ( ) 表示不包含端点，方括号 [ ] 表示包含端点。对于多个区间组成的解，使用并集符号 ∪，例如 (−∞, 1) ∪ (4, ∞)。
Be careful with infinity (∞): it is always accompanied by a parenthesis because infinity is not a number that can be reached. Also, when the solution is all real numbers, write ℝ or (−∞, ∞).
无穷大 ∞ 总是搭配圆括号使用，因为无穷大不是某个可以取到的具体数值。若解集为所有实数，则写作 ℝ 或 (−∞, ∞)。
10. Discriminant and Inequalities for Roots | 判别式与根的不等式
A common application involves using the discriminant to find conditions for a quadratic to have real and distinct roots, repeated roots, or no real roots. For ax² + bx + c = 0, the discriminant Δ = b² – 4ac. Set Δ > 0, Δ = 0, or Δ < 0 and solve the resulting inequality.
一个常见的应用是利用判别式找出一元二次方程具有不等实根、等根或无实根的条件。对于 ax² + bx + c = 0，判别式 Δ = b² – 4ac。令 Δ > 0、Δ = 0 或 Δ < 0，解相应的不等式即可。
For example, find k such that x² + kx + 4 = 0 has no real roots. The discriminant is k² – 16. Set k² – 16 < 0, giving –4 < k < 4. This type of question tests both your understanding of the discriminant and your ability to solve quadratic inequalities.
例如，求 k 的值使得 x² + kx + 4 = 0 无实根。判别式为 k² – 16。令 k² – 16 < 0，解得 –4 < k < 4。此类题目既考查对判别式的理解，也考查求解二次不等式的技能。
11. Inequalities in Context: Word Problems | 从实际问题中建立不等式
Many exam questions present a real-world scenario that requires you to formulate an inequality before solving. For instance, ‘A student scores p% on one paper and q% on another. To achieve a grade, the average must exceed 70%. Write an inequality and find possible scores.’ Model this as (p + q)/2 ≥ 70.
许多考试题目会给出实际问题情境，要求先建立不等式再进行求解。例如，“某生在一次测验中得 p%，在另一次中得 q%。要获得某个等级，平均分需超过 70%。写出不等式并求出可能的分数。”此时可建立不等式 (p + q)/2 ≥ 70。
Always define your variables clearly and interpret the final solution in the context of the problem. Rounding and discrete vs continuous data might affect the final answer, so read the question carefully.
务必清楚地定义变量，并在问题语境下解释最终的解。四舍五入以及数据是离散还是连续都会影响最终答案，因此要仔细审题。
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