2024年数学高考大一轮复习第十二章 §12.3　绝对值不等式（附答单独案解析）

§12.3　绝对值不等式

考试要求　1.理解绝对值的几何意义，并了解下列不等式成立的几何意义及取等号的条件：|a＋b|≤|a|＋|b|(a，b∈R)；|a－c|≤|a－b|＋|b－c|(a，b，c∈R).2.会利用绝对值的几何意义求解以下类型的不等式：|ax＋b|≤c，|ax＋b|≥c；|x－a|＋|x－b|≥c，|x－a|＋|x－b|≤c.

知识梳理

1．绝对值不等式的解法

(1)含绝对值的不等式|x|<a与|x|>a的解集

(2)|ax＋b|≤c(c>0)和|ax＋b|≥c(c>0)型不等式的解法

①|ax＋b|≤c⇔________________________.

②|ax＋b|≥c⇔________________________.

(3)|x－a|＋|x－b|≥c(c>0)和|x－a|＋|x－b|≤c(c>0)型不等式的解法

①利用绝对值不等式的几何意义求解，体现了数形结合的思想．

②利用“零点分段法”求解，体现了分类讨论的思想．

③通过构造函数，利用函数的图象求解，体现了函数与方程的思想．

2．含有绝对值的不等式的性质

(1)如果a，b是实数，则________≤|a±b|≤________.

(2)如果a，b，c是实数，那么______________，当且仅当________________时，等号成立．

思考辨析

判断下列结论是否正确(请在括号中打“√”或“×”)

(1)若|x|>c的解集为R，则c≤0.(　　)

(2)不等式|x－1|＋|x＋2|<2的解集为∅.(　　)

(3)对|a＋b|≥|a|－|b|当且仅当a>b>0时等号成立．(　　)

(4)对|a－b|≤|a|＋|b|当且仅当ab≤0时等号成立．(　　)

教材改编题

1．不等式3≤|5－2x|<9的解集为(　　)

A．[－2,1)∪[4,7)   B．(－2,1]∪(4,7]

C．(－2，－1]∪[4,7)   D．(－2,1]∪[4,7)

2．不等式|x－1|－|x－5|<2的解集为________．

3．设a，b∈R，|a－b|>2，则关于实数x的不等式|x－a|＋|x－b|>2的解集是________．

题型一　绝对值不等式的解法

例1 (2021·全国乙卷)已知函数f(x)＝|x－a|＋|x＋3|.

(1)当a＝1时，求不等式f(x)≥6的解集；

(2)若f(x)>－a，求a的取值范围．

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思维升华 解绝对值不等式的基本方法

(1)利用绝对值的定义，通过分类讨论转化为解不含绝对值符号的普通不等式．

(2)当不等式两端均为正数时，可通过两边平方的方法，转化为不含绝对值符号的普通不等式．

(3)利用绝对值的几何意义，数形结合求解．

跟踪训练1 已知函数f(x)＝－.

(1)画出函数y＝f(x)的图象；

(2)解不等式≥1.

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题型二　利用绝对值不等式的性质求最值

例2 已知函数f(x)＝|2x＋1|＋|x－4|.

(1)解不等式f(x)≤6；

(2)若不等式f(x)＋|x－4|<a2－8a有解，求实数a的取值范围．

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思维升华 求含绝对值函数的最值时，常用的方法有三种

(1)利用绝对值的几何意义．

(2)利用绝对值的三角不等式，即|a|＋|b|≥|a±b|≥||a|－|b||.

(3)利用零点分段法，转化为分段函数求最值．

跟踪训练2 已知函数f(x)＝－，m∈R.

(1)若m＝3，求不等式f(x)>1的解集；

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(2)若对∀x∈R，不等式f(x)＋2≥4都成立，求实数m的取值范围．

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题型三　绝对值不等式的综合应用

例3 设函数f(x)＝|2x＋1|＋|x－1|.

(1)画出y＝f(x)的图象；

(2)当x∈[0，＋∞)时，f(x)≤ax＋b，求a＋b的最小值．

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思维升华 (1)解决与绝对值有关的综合问题的关键是去掉绝对值，化为分段函数．

(2)数形结合是解决与绝对值有关的综合问题的常用方法．

跟踪训练3 (2023·成都联考)已知函数f(x)＝|x－2|－a|x＋1|.

(1)当a＝1时，求不等式f(x)<x的解集；

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(2)当a＝2时，若关于x的不等式f(x)>m＋1恰有2个整数解，求实数m的取值范围．

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