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}$