Bất phương trình bậc nhất 1 ẩn dạng chứa | |

Ta có:

$\left| {1 - 4x} \right| \ge 2x + 1 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 1 - 4x \ge 0\\ 1 - 4x \ge 2x + 1 \end{array} \right.\\ \left\{ \begin{array}{l} 1 - 4x < 0\\ 4x - 1 \ge 2x + 1 \end{array} \right. \end{array} \right.$

$\Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \le \frac{1}{4}\\ x \le 0 \end{array} \right.\\ \left\{ \begin{array}{l} x > \frac{1}{4}\\ x \ge 1 \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} x \le 0\\ x \ge 1 \end{array} \right.$

Ta có:

$\begin{align} & \left| x+2 \right| > \left| x-3 \right|\Leftrightarrow {{\left( x+2 \right)}^ 2 } > {{\left( x-3 \right)}^ 2 } \\ & \Leftrightarrow 2 x -1 > 0\Rightarrow x > \dfrac{1}{2} \\ \end{align}$

Ta có

$\left| 2x-1 \right| < \left| x-2 \right|\\\Leftrightarrow {{\left( 2x-1 \right)}^ 2 } < {{\left( x-2 \right)}^ 2 } \\ \Leftrightarrow 3{ x ^ 2 }-3 < 0\Leftrightarrow-1 < x < 1. \\ $

Tập nghiệm của bất phương trình là $x \in (-1;1).$

Vậy bất phương trình có một nghiệm nguyên là $x=0.$

Ta có : $\left| {3 - 5x} \right| \le x + 2 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 3 - 5x \ge 0\\ 3 - 5x \le x + 2 \end{array} \right.\\ \left\{ \begin{array}{l} 3 - 5x < 0\\ - 3 + 5x \le x + 2 \end{array} \right. \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} \frac{1}{6} \le x \le \frac{3}{5}\\ \frac{3}{5} < x \le \frac{5}{4} \end{array} \right. \Leftrightarrow \frac{1}{6} \le x \le \frac{5}{4}$

Ta có 

$\left| {6x + 1} \right| < 3 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 6x + 1 \ge 0\\ 6x + 1 < 3 \end{array} \right.\\ \left\{ \begin{array}{l} 6x + 1 < 0\\  - 6x - 1 < 3 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \dfrac{{ - 1}}{6}\\ x < \dfrac{1}{3} \end{array} \right.\\ \left\{ \begin{array}{l} x < \dfrac{{ - 1}}{6}\\ x > \dfrac{{ - 2}}{3} \end{array} \right. \end{array} \right. \Leftrightarrow  - \dfrac{2}{3} < x < \dfrac{1}{3}.$

Vậy tập nghiệm của bất phương trình là $\left( { - \dfrac{2}{3};\dfrac{1}{3}} \right)$.

TH1: $x > 3$

$\dfrac{2x-5}{\left| x-3 \right|}\ge 3\Leftrightarrow \dfrac{2x-5}{x-3}\ge 3\Leftrightarrow \dfrac{4-x}{x-3}\ge 0\Leftrightarrow 3 < x\le 4$

Kết hợp điều kiện : $3 < x\le 4$

TH2: $x\le 3$

$\dfrac{2x-5}{\left| x-3 \right|}\ge 3\Leftrightarrow \dfrac{2x-5}{-(x-3)}\ge 3\Leftrightarrow \dfrac{5x-14}{3-x}\le 0\Leftrightarrow \dfrac{14} 5 \le x < 3$

Kết hợp điều kiện: $x\in \left[ \dfrac{14} 5 ;3 \right)$

KL: $x\in \left[ \dfrac{14} 5 ;4 \right]\backslash \left\{ 3 \right\}$

TH1:

$\dfrac{x+1}{3x}\ge 0\Leftrightarrow \left[ \begin{align} & x > 0 \\ & x\le -1 \\ \end{align} \right.$

$\Rightarrow \dfrac{x+1}{3x} < 1\Leftrightarrow \dfrac{1-2x}{3x} < 0\Leftrightarrow \left[ \begin{align} & x < 0 \\ & x > \dfrac{1}{2} \\ \end{align} \right.$

Kết hợp điều kiện ta có: $\left[ \begin{align} & x\le -1 \\ & x > \dfrac{1}{2} \\ \end{align} \right.$

TH2:

$\dfrac{x+1}{3 x } < 0\Leftrightarrow -1 < x < 0$

$\Rightarrow -\dfrac{x+1}{3x} < 1\Leftrightarrow \dfrac{4x+1}{3x} > 0\Leftrightarrow \left[ \begin{align} & x > 0 \\ & x < -\dfrac{1}{4} \\ \end{align} \right.$

Kết hợp điều kiện : $-1 < x < -\dfrac{1}{4}$

Vậy BPT có nghiệm là $\left[ \begin{align} & x < -\dfrac{1}{4} \\ & x > \dfrac{1}{2} \\ \end{align} \right.$

Ta có : $\left| {3x - 1} \right| > x \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 3x - 1 \ge 0\\ 3x - 1 > x \end{array} \right.\\ \left\{ \begin{array}{l} 3x - 1 < 0\\ 3x - 1 < - x \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x > \frac{1}{2}\\ x < \frac{1}{4} \end{array} \right.$

Ta có:  $\left| {4x - 9} \right| > 5x + 1 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 4x - 9 \ge 0\\ 4x - 9 > 5x + 1 \end{array} \right.\\ \left\{ \begin{array}{l} 4x - 9 < 0\\ - 4x + 9 > 5x + 1 \end{array} \right. \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \frac{9}{4}\\ x < - 10 \end{array} \right.\\ \left\{ \begin{array}{l} x < \frac{9}{4}\\ x < \frac{8}{9} \end{array} \right. \end{array} \right. \Leftrightarrow x < \frac{8}{9}$

Vậy BPT không có nghiệm nguyên dương.

Ta có:

$\left| {5 - 3x} \right| \le x + 3 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 5 - 3x \ge 0\\ 5 - 3x \le x + 3 \end{array} \right.\\ \left\{ \begin{array}{l} 5 - 3x < 0\\ 3x - 5 \le x + 3 \end{array} \right. \end{array} \right.$

$\Rightarrow \left[ {\begin{array}{*{20}{l}} {\left\{ {\begin{array}{*{20}{l}} {x \leqslant \frac{5}{3}} \\ {x \geqslant \frac{1}{2}} \end{array}} \right.} \\ {\left\{ {\begin{array}{*{20}{l}} {x > \frac{5}{3}} \\ {x \leqslant 4} \end{array}} \right.} \end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}} {x \in \left[ {\frac{1}{2};\frac{5}{3}} \right]} \\ {x \in \left( {\frac{5}{3};4} \right]} \end{array}} \right. \Rightarrow x \in \left[ {\frac{1}{2};4} \right]$

Vậy BPT có các nghiệm nguyên là $1;2;3;4$

Ta có:

$\left| {7 - 2x} \right| < 3x \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 7 - 2x \ge 0\\ 7 - 2x < 3x \end{array} \right.\\ \left\{ \begin{array}{l} 7 - 2x < 0\\ 2x - 7 < 3x \end{array} \right. \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \le \dfrac{7}{2}\\ x > \dfrac{7}{5} \end{array} \right.\\ \left\{ \begin{array}{l} x > \dfrac{7}{2}\\ x >  - 7 \end{array} \right. \end{array} \right. \Leftrightarrow x > \dfrac{7}{5}$

TH1:

$\dfrac{x+1}{2x-1}\ge 0\Leftrightarrow \left[ \begin{align} & x\le -1 \\ & x > \dfrac{1}{2} \\ \end{align} \right.$

$\Rightarrow \dfrac{x+1}{2x-1}\le 2\Leftrightarrow \dfrac{3x-3}{2x-1}\ge 0\Leftrightarrow \left[ \begin{align} & x\ge 1 \\ & x < \dfrac{1}{2} \\ \end{align} \right.$

Kết hợp điều kiện ta có: $\left[ \begin{align} & x\ge 1 \\ & x\le -1 \\ \end{align} \right.$

TH2:

$\dfrac{x+1}{2x-1} < 0\Leftrightarrow -1 < x < \dfrac{1}{2}$

$\Rightarrow -\dfrac{x+1}{2x-1}\le 2\Leftrightarrow \dfrac{5x-1}{1-2x}\le 0\Leftrightarrow \left[ \begin{align} & x > \dfrac{1}{2} \\ & x\le \dfrac{1}{5} \\ \end{align} \right.$

Kết hợp điều kiện : $-1 < x\le \dfrac{1}{5}$

KL: $x\in \left( -\infty ;\dfrac{1}{5} \right]\cup \left[ 1;+\infty \right)$

Ta có: 

$\left| {x + 4} \right| \le - 2x + 5 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x + 4 \ge 0\\ x + 4 \le - 2x + 5 \end{array} \right.\\ \left\{ \begin{array}{l} x + 4 < 0\\ - x - 4 \le - 2x + 5 \end{array} \right. \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge - 4\\ x \le \frac{1}{3} \end{array} \right.\\ \left\{ \begin{array}{l} x < - 4\\ x \le 3 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 4 \le x \le \frac{1}{3}\\ x < - 4 \end{array} \right. \Leftrightarrow x \le \frac{1}{3}$

Vậy $a=1,b=3\Rightarrow { a ^ 2 }-{ b ^ 2 }=-8$

TH1: $\dfrac{3x+4}{x-2}\ge 0\Leftrightarrow \left[ \begin{align} & x > 2 \\ & x\le -\dfrac{4}{3} \\ \end{align} \right.$

$\left| \dfrac{3x+4}{x-2} \right|\le 3\Leftrightarrow \dfrac{3x+4}{x-2}\le 3\Leftrightarrow \dfrac{10}{x-2}\le 0\Leftrightarrow x-2 < 0\Leftrightarrow x < 2$

Kết hợp điều kiện : $x\le -\dfrac{4}{3}$

TH2: $\dfrac{3x+4}{x-2} < 0\Leftrightarrow -\dfrac{4}{3} < x < 2$

$\left| \dfrac{3x+4}{x-2} \right|\le 3\Leftrightarrow -\dfrac{3x+4}{x-2}\le 3\Leftrightarrow \dfrac{6x-2}{2-x}\le 0\Leftrightarrow \left[ \begin{align} & x > 2 \\ & x\le \dfrac{1}{3} \\ \end{align} \right.$

Kết hợp điều kiện: $-\dfrac{4}{3} \le x\le \dfrac{1}{3}$

KL : $x\le \dfrac{1}{3}$

Ta có:

$\left| {2x - 5} \right| \le x + 1 \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}2x - 5 \ge 0\\2x - 5 \le x + 1\end{array} \right.\\\left\{ \begin{array}{l}2x - 5 < 0\\5 - 2x \le x + 1\end{array} \right.\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge \frac{5}{2}\\x \le 6\end{array} \right.\\\left\{ \begin{array}{l}x < \frac{5}{2}\\x \ge \frac{4}{3}\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}x \in \left[ {\frac{5}{2};6} \right]\\x \in \left[ {\frac{4}{3};\frac{5}{2}} \right)\end{array} \right. \Rightarrow x \in \left[ {\frac{4}{3};6} \right]$

Vậy các nghiệm nguyên là 2; 3; 4; 5; 6 suy ra  tổng các nghiệm nguyên của BPT là $20$

Ta có : $\left| {2x + 3} \right| \ge x + 1 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 2x + 3 \ge 0\\ 2x + 3 \ge x + 1 \end{array} \right.\\ \left\{ \begin{array}{l} 2x + 3 < 0\\ - 2x - 3 \ge x + 1 \end{array} \right. \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \dfrac{{ - 3}}{2}\\ x \ge - 2 \end{array} \right.\\ \left\{ \begin{array}{l} x < \dfrac{{ - 3}}{2}\\ x \le \dfrac{{ - 4}}{3} \end{array} \right. \end{array} \right. \Rightarrow x \in R$

TH1: $2-5x\ge 0\Leftrightarrow x\le \dfrac{2}{5}$.

Bất phương trình có dạng: $2-5 x \ge x+1\Leftrightarrow 6 x \le 1\Leftrightarrow x\le \dfrac{1}{6}$.

TH2: $2-5x < 0\Leftrightarrow x > \dfrac{2}{5}$.

Bất phương trình có dạng: $5 x -2\ge x+1\Leftrightarrow 4 x \ge 3\Leftrightarrow x\ge \dfrac{3}{4}$.

Kết hợp với điều kiện ta có $x\in \left( -\infty ;\dfrac{1}{6} \right]\cup \left[ \dfrac{3}{4} ;+\infty \right)$.

Ta có : $\left| {x + 1} \right| \ge 3 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge  - 1\\ x \ge 2 \end{array} \right.\\ \left\{ \begin{array}{l} x <  - 1\\ x \le  - 4 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x \ge 2\\ x \le  - 4 \end{array} \right. \Rightarrow x \in \left( { - \infty ; - 4} \right] \cup \left[ {2; + \infty } \right)$

Ta có: $\left| {2 - x} \right| > 4 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \le 2\\ x <  - 2 \end{array} \right.\\ \left\{ \begin{array}{l} x > 2\\ x > 6 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x <  - 2\\ x > 6 \end{array} \right. \Rightarrow x \in \left( { - \infty ; - 2} \right) \cup \left( {6; + \infty } \right)$

KL: $x\in \left( -\infty ;-4 \right)\cup \left( 6;+\infty \right)$

TH1:

$\dfrac{x-1}{2x+1}\ge 0\Leftrightarrow \left[ \begin{align} & x\ge 1 \\ & x < -\dfrac{1}{2} \\ \end{align} \right.$

$\Rightarrow \dfrac{x-1}{2x+1} < 3\Leftrightarrow \dfrac{5x+4}{2x+1} > 0\Leftrightarrow \left[ \begin{align} & x < -\dfrac{4}{5} \\ & x > -\dfrac{1}{2} \\ \end{align} \right.$

Kết hợp điều kiện ta có : $\left[ \begin{align} & x\ge 1 \\ & x < -\dfrac{4}{5} \\ \end{align} \right.$

TH2:

$\dfrac{x-1}{2x+1} < 0\Leftrightarrow -\dfrac{1}{2} < x < 1$

$\Rightarrow -\dfrac{x-1}{2x+1} < 1\Leftrightarrow \dfrac{3x}{2x+1} > 0\Leftrightarrow \left[ \begin{align} & x > 0 \\ & x < -\dfrac{1}{2} \\ \end{align} \right.$

Kết hợp điều kiện ta có : $0 < x < 1$

KL: $x\in \left( -\infty ;-\dfrac{4}{5} \right)\cup \left( 0;+\infty \right)$

Ta có 

$\left| {2x + 5} \right| \ge 4x + 7\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 2x + 5 \ge 0\\ 2x + 5 \ge 4x + 7 \end{array} \right.\\ \left\{ \begin{array}{l} 2x + 5 < 0\\ - 2x - 5 \ge 4x + 7 \end{array} \right. \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge - \dfrac{5}{2}\\ x \le - 1 \end{array} \right.\\ \left\{ \begin{array}{l} x < - \dfrac{5}{2}\\ x \le - 2 \end{array} \right. \end{array} \right. \\ \Leftrightarrow \left[ \begin{array}{l} - \dfrac{5}{2} \le x \le - 1\\ x < - \dfrac{5}{2} \end{array} \right.\\ \Leftrightarrow x \le - 1.$.

Vậy BPT có vô số nghiệm nguyên.

TH1: $x > 3$

$\dfrac{2x-5}{\left| x-3 \right|}+1 > 2\Leftrightarrow \dfrac{2x-5}{x-3}-1 > 0\Leftrightarrow \dfrac{x-2}{x-3} > 0\Leftrightarrow \left[ \begin{align} & x > 3 \\ & x < 2 \\ \end{align} \right.$

Kết hợp với điều kiện ta có $x > 3$

TH2: $x\le 3$

$\dfrac{2x-5}{\left| x-3 \right|}+1 > 2\Leftrightarrow \dfrac{2x-5}{-(x-3)}-1 > 0\Leftrightarrow \dfrac{3x-8}{3-x} > 0\Leftrightarrow \dfrac{8}{3} < x < 3$

KL: $x\in \left( \dfrac{8}{3} ;3 \right)\cup \left( 3;+\infty \right)$

Ta có:

$\left| {7x - 9} \right| > 4x + 6 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 7x - 9 \ge 0\\ 7x - 9 > 4x + 6 \end{array} \right.\\ \left\{ \begin{array}{l} 7x - 9 < 0\\ 9 - 7x > 4x + 6 \end{array} \right. \end{array} \right.$ $\Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \frac{9}{7}\\ x > 5 \end{array} \right.\\ \left\{ \begin{array}{l} x < \frac{9}{7}\\ x < \frac{3}{{11}} \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} x > 5\\ x < \frac{3}{{11}} \end{array} \right.$

$|2x+1| < 3x\Leftrightarrow -3x < 2x+1 < 3x\Leftrightarrow \left\{ \begin{array}{l} & x > \dfrac{-1}{5} \\ & x > 1 \\ \end{array} \right.\Leftrightarrow x > 1$
Suy ra giá trị nguyên $x$ trong $[-2019;2019]$ thỏa mãn bất phương trình $|2x+1| < 3x$ là
$\text{ }\!\!\{\!\!\text{ 2;3;}....\text{2019 }\!\!\}\!\!\text{ }$ gồm 2018 số.

Ta có: 

$\left| {3 - 4x} \right| \le x - 2 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 3 - 4x \ge 0\\ 3 - 4x \le x - 2 \end{array} \right.\\ \left\{ \begin{array}{l} 3 - 4x < 0\\ - 3 + 4x \le x - 2 \end{array} \right. \end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \le \frac{3}{4}\\ x \ge 1 \end{array} \right.\\ \left\{ \begin{array}{l} x > \frac{3}{4}\\ x \le \frac{1}{3} \end{array} \right. \end{array} \right.\left( {VN} \right)$

TH1: $x > 3$

$\dfrac{2 x -5}{\left| x-3 \right|}+1 > 0\Leftrightarrow \dfrac{2 x -5}{x-3}+1 > 0\Leftrightarrow \dfrac{3 x -8}{x-3} > 0\Leftrightarrow \left[ \begin{align} & x < \dfrac{8}{3} \\ & x > 3 \\ \end{align} \right.$

Kết hợp điều kiện ta có $x > 3$

TH2: $x < 3$

$\dfrac{2 x -5}{\left| x-3 \right|}+1 > 0\Leftrightarrow \dfrac{2 x -5}{-(x-3)}+1 > 0\Leftrightarrow \dfrac{2-x}{x-3} > 0\Leftrightarrow 2 < x < 3$ (thỏa mãn điều kiện)

KL: $x\in \left( 2;+\infty \right)\backslash \left\{ 3 \right\}$

Ta có

$\left| {3x - 5} \right| \ge 2x - 1 \\\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 3x - 5 \ge 0\\ 3x - 5 \ge 2x - 1 \end{array} \right.\\ \left\{ \begin{array}{l} 3x - 5 < 0\\ - 3x + 5 \ge 2x - 1 \end{array} \right. \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \dfrac{5}{3}\\ x \ge 4 \end{array} \right.\\ \left\{ \begin{array}{l} x < \dfrac{5}{3}\\ x \le \dfrac{6}{5} \end{array} \right. \end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l} x \ge 4\\ x \le \dfrac{6}{5}. \end{array} \right.$

Vậy BPT có tập nghiệm $\left( -\infty ;\dfrac{6}{5} \right]\cup \left[ 4;+\infty \right)\Rightarrow a=\dfrac{6}{5} ;b=4\Rightarrow a-b=-\dfrac{14} 5$.

Ta có : $\left| {x + 1} \right| \ge 3x - 2 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge  - 1\\ x \le \frac{3}{2} \end{array} \right.\\ \left\{ \begin{array}{l} x <  - 1\\ x \le \frac{1}{4} \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}  - 1 \le x \le \frac{3}{2}\\ x <  - 1 \end{array} \right. \Leftrightarrow x \le \frac{3}{2}$

Ta có: $\left| {2 - x} \right| > 2x + 1 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \le 2\\ x < \frac{1}{3} \end{array} \right.\\ \left\{ \begin{array}{l} x > 2\\ x <  - 3 \end{array} \right. \end{array} \right. \Leftrightarrow x < \frac{1}{3}$

KL: $x\in \left( -\infty ;\dfrac{1}{3} \right)$

Ta có:

$\left| {2 - 5x} \right| \ge x + 1 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 2 - 5x \ge 0\\ 2 - 5x \ge x + 1 \end{array} \right.\\ \left\{ \begin{array}{l} 2 - 5x < 0\\ 5x - 2 \ge x + 1 \end{array} \right. \end{array} \right.$

$\Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \le \frac{2}{5}\\ x \le \frac{1}{6} \end{array} \right.\\ \left\{ \begin{array}{l} x > \frac{2}{5}\\ x \ge \frac{3}{4} \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} x \le \frac{1}{6}\\ x \ge \frac{3}{4} \end{array} \right.$

Ta thấy BPT có vô số nghiệm nguyên trong đó có nghiệm $0$ nên tích các nghiệm nguyên của BPT bằng $0$

TH1: $x < -\dfrac{1}{2} \Rightarrow -2x-1\le 3-x\Leftrightarrow x\ge -4$ . Kết hợp điều kiện $-4\le x < -\dfrac{1}{2}$

TH2: $-\dfrac{1}{2} \le x\le 3\Rightarrow 2x+1\le 3-x\Leftrightarrow x\le \dfrac{2}{3}$ . Kết hợp điều kiện $-\dfrac{1}{2} \le x\le \dfrac{2}{3}$

TH3: $x > 3\Rightarrow 2x+1\le x-3\Leftrightarrow x\le -4$ . Kết hợp điều kiện: vô nghiệm

KL: $-4\le x < \dfrac{2}{3}$

Ta có : $2\left( \left| x \right|+1 \right)\ge 4\left| x \right|+9$ (1)

Đặt $t=\left| x \right|\left( t\ge 0 \right)$ . Từ (1) $\Rightarrow 2\left( t+1 \right)\ge 4t+9\Leftrightarrow t\le -\dfrac{7}{2}$ (vô lí)

Vậy BPT vô nghiệm

Ta có: 

$$\left| {7x - 3} \right| < 2x + 1 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 7x - 3 \ge 0\\ 7x - 3 < 2x + 1 \end{array} \right.\\ \left\{ \begin{array}{l} 7x - 3 < 0\\ - 7x + 3 < 2x + 1 \end{array} \right. \end{array} \right.$$ $\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \dfrac{3}{7}\\ x < \dfrac{4}{5} \end{array} \right.\\ \left\{ \begin{array}{l} x < \dfrac{3}{7}\\ x > \dfrac{2}{9} \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \dfrac{3}{7} \le x < \dfrac{4}{5}\\ \dfrac{2}{9} < x < \dfrac{3}{7} \end{array} \right. \Leftrightarrow \dfrac{2}{9} < x < \dfrac{4}{5}$