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Solving Absolute Value Equations and Inequalities | 绝对值方程与不等式求解技巧

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📚 Solving Absolute Value Equations and Inequalities | 绝对值方程与不等式求解技巧

Absolute value is one of the most frequently tested topics in A-Level mathematics because it combines algebraic manipulation with geometric intuition. Mastery of absolute value equations and inequalities requires a clear understanding of the piecewise definition, the distance interpretation, and careful handling of signs. This guide presents the essential techniques, common pitfalls, and exam-style examples to help you solve these problems accurately and efficiently.

绝对值是A-Level数学中的高频考点，因为它将代数运算与几何直观紧密结合。要熟练求解绝对值方程与不等式，你需要清晰理解分段定义、数轴距离的意义，并谨慎处理符号。本指南将系统讲解核心技巧、常见错误以及真题型例题，帮助你准确、高效地解决这类问题。

1. Definition and Core Properties of Absolute Value | 绝对值的定义与核心性质

The absolute value of a real number x, written |x|, is its distance from zero on the number line. It is always non-negative. The formal definition is piecewise: |x| = x when x ≥ 0, and |x| = −x when x < 0.

实数 x 的绝对值，记作 |x|，表示它在数轴上到原点的距离，因此永远非负。其严格定义为分段形式：当 x ≥ 0 时，|x| = x；当 x < 0 时，|x| = −x。

  • |a| ≥ 0, and |a| = 0 if and only if a = 0.

    |a| ≥ 0，且 |a| = 0 当且仅当 a = 0。

  • |ab| = |a| × |b| and |a / b| = |a| / |b| when b ≠ 0.

    |ab| = |a| × |b|，且当 b ≠ 0 时，|a / b| = |a| / |b|。

  • |a|² = a², which is often used to eliminate absolute value signs.

    |a|² = a²，这一性质常用于消去绝对值符号。

  • Triangle inequality: |a + b| ≤ |a| + |b|.

    三角不等式：|a + b| ≤ |a| + |b|。

2. Basic Absolute Value Equations | 基础绝对值方程

For an equation of the form |f(x)| = k, where k ≥ 0, the solution is found by removing the absolute value and writing two separate equations: f(x) = k or f(x) = −k. If k < 0, there is no solution because an absolute value can never be negative.

对于形如 |f(x)| = k 的方程，其中 k ≥ 0，求解时去掉绝对值并写出两个独立方程：f(x) = k 或 f(x) = −k。若 k < 0，则无解，因为绝对值不可能为负数。

|f(x)| = k (k ≥ 0) ⇔ f(x) = k or f(x) = −k

Example: Solve |2x − 3| = 7. Then 2x − 3 = 7 gives x = 5, and 2x − 3 = −7 gives x = −2. The solution set is x = 5 or x = −2.

例：解 |2x − 3| = 7。由 2x − 3 = 7 得 x = 5；由 2x − 3 = −7 得 x = −2。解集为 x = 5 或 x = −2。

Always check your answers in the original equation, especially when the right-hand side involves a variable.

务必把答案代回原方程检验，尤其是当右侧含有变量时。

3. Equations with a Variable Right-Hand Side | 含变量右侧的绝对值方程

When the equation has the form |f(x)| = g(x), the right-hand side must be non-negative for any solution to exist. You can first impose the condition g(x) ≥ 0, then solve f(x) = g(x) or f(x) = −g(x). Finally, verify each candidate solution against the original equation.

当方程为 |f(x)| = g(x) 时，右侧 g(x) 必须非负才可能有解。可以先限定 g(x) ≥ 0，再解 f(x) = g(x) 或 f(x) = −g(x)，最后将每个候选解代回原方程验证。

Example: Solve |x − 1| = 2x − 1. First note 2x − 1 ≥ 0, so x ≥ 1/2. Then x − 1 = 2x − 1 gives x = 0, which fails the condition. Or x − 1 = −(2x − 1) gives x − 1 = −2x + 1, so 3x = 2, x = 2/3. Since 2/3 ≥ 1/2, the only solution is x = 2/3.

例：解 |x − 1| = 2x − 1。首先 2x − 1 ≥ 0，因此 x ≥ 1/2。由 x − 1 = 2x − 1 得 x = 0，不满足条件；由 x − 1 = −(2x − 1) 得 x − 1 = −2x + 1，即 3x = 2，x = 2/3。因为 2/3 ≥ 1/2，所以唯一解为 x = 2/3。

Alternatively, square both sides: (x − 1)² = (2x − 1)², solve the resulting quadratic, and reject extraneous roots by substitution.

另一种方法是两边平方：(x − 1)² = (2x − 1)²，解出二次方程后再代入原方程排除增根。

4. Absolute Value as Distance on the Number Line | 绝对值的数轴距离意义

The expression |x − a| represents the distance between x and a on the number line. This interpretation is extremely useful for solving inequalities and understanding solution sets geometrically.

表达式 |x − a| 表示数轴上 x 与 a 之间的距离。这种解释对解不等式和直观理解解集非常有用。

For |x − a| = k, the solutions are the two points located exactly k units from a: x = a − k and x = a + k.

对于 |x − a| = k，解是距离 a 恰好 k 个单位的两个点：x = a − k 与 x = a + k。

For |x − a| < k, the solution is the open interval (a − k, a + k). For |x − a| > k, the solution is the union of two rays: (−∞, a − k) ∪ (a + k, ∞).

对于 |x − a| < k，解集为开区间 (a − k, a + k)。对于 |x − a| > k，解集为两个射线之并集：(−∞, a − k) ∪ (a + k, ∞)。

|x − a| < k ⇔ a − k < x < a + k
|x − a| > k ⇔ x < a − k or x > a + k

You should draw a number line for simple cases; picture the distance and mark the intervals correctly.

遇到简单情况时建议画数轴，将距离可视化并正确标记区间。

5. “Less Than” Inequalities |x| < k | 小于型不等式 |x| < k

For k > 0, the inequality |f(x)| < k is equivalent to −k < f(x) < k, which is a pair of simultaneous inequalities: f(x) > −k and f(x) < k. Remember to reverse the inequality when you multiply or divide by a negative number.

当 k > 0 时，不等式 |f(x)| < k 等价于 −k < f(x) < k，即两个联立不等式：f(x) > −k 且 f(x) < k。注意在乘以或除以负数时要把不等号方向反过来。

Example: Solve |x − 2| < 3. Then −3 < x − 2 < 3, so add 2 throughout: −1 < x < 5. The solution set is the open interval (−1, 5).

例：解 |x − 2| < 3。则 −3 < x − 2 < 3，各项同时加 2 得 −1 < x < 5。解集为开区间 (−1, 5)。

If the inequality is ≤, the endpoints are included, so the solution is a closed interval [a, b]. Always match the original inequality symbol when writing your final answer.

若不等式为 ≤，则端点包含在内，解集为闭区间 [a, b]。书写最终答案时，务必与原不等式符号保持一致。

6. “Greater Than” Inequalities |x| > k | 大于型不等式 |x| > k

For k > 0, the inequality |f(x)| > k is equivalent to f(x) > k or f(x) < −k. The word “or” is essential because the solution set is a union of two disjoint regions.

当 k > 0 时，不等式 |f(x)| > k 等价于 f(x) > k 或 f(x) < −k。这里“或”字至关重要，因为解集是两个互不相交区域的并集。

Example: Solve |2x + 1| > 5. Then 2x + 1 > 5 or 2x + 1 < −5. The first gives x > 2, and the second gives x < −3. Therefore the solution is x < −3 or x > 2, written as (−∞, −3) ∪ (2, ∞).

例：解 |2x + 1| > 5。则 2x + 1 > 5 或 2x + 1 < −5。前者得 x > 2，后者得 x < −3。因此解为 x < −3 或 x > 2，即 (−∞, −3) ∪ (2, ∞)。

Inequality 不等式 Equivalence 等价形式 Solution set 解集
|f(x)| < k −k < f(x) < k Interval 区间
|f(x)| > k f(x) > k or f(x) < −k Union of two intervals 两个区间之并

A common mistake is to write −k < f(x) < k for a “greater than” inequality. This is incorrect; “greater than” produces two separate branches, not a single bounded interval.

常见错误是把“大于型”不等式写成 −k < f(x) < k。这是不对的；大于型会产生两个独立分支，而不是单一有界区间。

7. Method of Critical Points for Multiple Absolute Values | 零点分段法处理多个绝对值

When an equation or inequality contains more than one absolute value term, use the critical points method. First find the values of x that make each expression inside absolute value bars equal to zero. These critical points split the real number line into intervals. On each interval, determine the sign of each expression and remove the absolute value bars accordingly.

当方程或不等式中含有多于一个绝对值项时，使用零点分段法。首先求出使每个绝对值内部表达式等于零的 x 值，这些零点把实数轴分成若干区间。在每个区间上判断各表达式的符号，并据此去掉绝对值符号。

Example: Solve |x − 1| + |x − 3| = 4. Critical points are x = 1 and x = 3. Consider three intervals.

例：解 |x − 1| + |x − 3| = 4。零点为 x = 1 和 x = 3。分三个区间讨论。

  • For x < 1: −(x − 1) − (x − 3) = 4, so −2x + 4 = 4, giving x = 0, which lies in this interval.

    当 x < 1 时：−(x − 1) − (x − 3) = 4，得 −2x + 4 = 4，即 x = 0，落在该区间内。

  • For 1 ≤ x ≤ 3: (x − 1) − (x − 3) = 2, which never equals 4, so no solution here.

    当 1 ≤ x ≤ 3 时：(x − 1) − (x − 3) = 2，不可能等于 4，因此该区间无解。

  • For x > 3: (x − 1) + (x − 3) = 4, so 2x − 4 = 4, giving x = 4, which lies in this interval.

    当 x > 3 时：(x − 1) + (x − 3) = 4，得 2x − 4 = 4，即 x = 4，落在该区间内。

Thus the solutions are x = 0 and x = 4. Always state which interval each solution belongs to when doing this method.

因此解为 x = 0 与 x = 4。用这种方法时，要明确每个解属于哪个区间。

8. Squaring and Substitution Techniques | 平方与换元技巧

Squaring both sides can remove absolute value bars, because |a|² = a². However, squaring may introduce extraneous solutions, so you must check every candidate in the original equation.

两边平方可以去掉绝对值，因为 |a|² = a²。但平方可能产生增根，所以必须将每个候选解代回原方程检验。

Example: Solve |x − 1| = 2x − 1. Squaring gives (x − 1)² = (2x − 1)², which expands to x² − 2x + 1 = 4x² − 4x + 1, so 3x² − 2x = 0, hence x = 0 or x = 2/3. Substituting x = 0 gives |−1| = 1 but 2(0) − 1 = −1, so reject 0. The valid solution is x = 2/3.

例：解 |x − 1| = 2x − 1。两边平方得 (x − 1)² = (2x − 1)²，展开为 x² − 2x + 1 = 4x² − 4x + 1，即 3x² − 2x = 0，所以 x = 0 或 x = 2/3。代回检验 x = 0 时 |−1| = 1，但 2(0) − 1 = −1，故舍去；有效解为 x = 2/3。

Substitution is useful when the equation is symmetric in |x|. For example, |x|² − 5|x| + 6 = 0 can be treated as a quadratic in |x|: let t = |x|, then t² − 5t + 6 = 0, so t = 2 or t = 3. Hence x = ±2 or x = ±3.

换元法适用于关于 |x| 对称的方程。例如 |x|² − 5|x| + 6 = 0 可看作关于 |x| 的二次方程：令 t = |x|，则 t² − 5t + 6 = 0，解得 t = 2 或 t = 3，因此 x = ±2 或 x = ±3。

9. Common Mistakes and Pitfalls | 常见错误与陷阱

Many students lose marks on absolute value questions because of avoidable sign errors. One common mistake is forgetting that if the right-hand side of an equation is negative, there is no solution. Another is confusing the “less than” and “greater than” inequality patterns.

很多学生在绝对值题目中丢分，是因为可以避免的符号错误。常见错误之一是忘记当方程右侧为负数时无解；另一个是混淆“小于型”和“大于型”不等式的模式。

  • For |f(x)| = g(x), you must check g(x) ≥ 0, not just solve two equations blindly.

    对于 |f(x)| = g(x)，必须检查 g(x) ≥ 0，不要盲目地拆成两个方程。

  • For |f(x)| > k, never write −k < f(x) < k; the correct form is f(x) > k or f(x) < −k.

    对于 |f(x)| > k，绝不写成 −k < f(x) < k；正确形式是 f(x) > k 或 f(x) < −k。

  • When squaring both sides, reject solutions that fail the original equation.

    两边平方后，必须舍去不满足原方程的解。

  • In the critical points method, do not forget to check that your solution lies within the interval you are considering.

    使用零点分段法时，不要忘记检查所得解是否落在你正在讨论的区间内。

  • Use the correct set notation: union ∪ for “or”, intersection ∩ for “and”.

    使用正确的集合符号：并集 ∪ 表示“或”，交集 ∩ 表示“且”。

10. Exam Tips and Worked Examples | 考场技巧与真题实战

In an exam, always sketch a number line or a simple graph when possible. For a single absolute value, the graph is a V-shape; inequalities can be solved by comparing the graph with a horizontal line. For multiple absolute values, the critical points method is systematic and reliable.

考试时，尽量画数轴或简单图像。单个绝对值的图像呈 V 形；解不等式时可将图像与水平线比较。对于多个绝对值，零点分段法最为系统可靠。

Worked example: Solve |x − 1| + |x − 3| ≥ 6.

真题演练：解 |x − 1| + |x − 3| ≥ 6。

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