2024年高考数学第一轮复习讲义第十三章13.3　绝对值不等式(学生版+解析)

这是一份2024年高考数学第一轮复习讲义第十三章13.3　绝对值不等式(学生版+解析)，共14页。

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|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|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)＝eq \b\lc\|\rc\|(\a\vs4\al\c1(x＋2))－eq \b\lc\|\rc\|(\a\vs4\al\c1(2x－3)).
(1)画出函数y＝f(x)的图象；
(2)解不等式eq \b\lc\|\rc\|(\a\vs4\al\c1(fx))≥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|1的解集；
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(2)若对∀x∈R，不等式f(x)＋2eq \b\lc\|\rc\|(\a\vs4\al\c1(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)m＋1恰有2个整数解，求实数m的取值范围．
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________________________________________________________________________不等式
a>0
a＝0
a0)和|ax＋b|≥c(c>0)型不等式的解法
①|ax＋b|≤c⇔－c≤ax＋b≤c.
②|ax＋b|≥c⇔ax＋b≥c或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||≤|a±b|≤|a|＋|b|.
(2)如果a，b，c是实数，那么|a－c|≤|a－b|＋|b－c|，当且仅当(a－b)(b－c)≥0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|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|