Giải thích các bước giải:

 2 

theo đề bài ta có x>0

a/

$4x-8<3x\\⇔x<8$

b/

$6x-12>3x\\⇔3x>12\\⇔x>4$

3

$|-5x+2|=\frac{1}{5}\\⇔\left \{ {{-5x+2=\frac{1}{5}} \atop {-5x+2=-\frac{1}{5}}} \right.\\⇔\left \{ {{x=\frac{9}{25}} \atop {x=\frac{11}{25}}} \right.$

b

$|3x+2|=-3\\⇔\left \{ {{3x+2=-3} \atop {3x+2=3}} \right.\\⇔\left \{ {{x=\frac{-5}{3}} \atop {x=\frac{1}{3}}} \right.$

C,D em xem trong hình

#X

Đáp án+Giải thích các bước giải:

 Bài 2:

\(\dfrac{x-2}{3x}<\dfrac{1}{4}\\(x\ne 0)\\\Leftrightarrow \dfrac{4(x-2)-3x}{12x}<0\\\Leftrightarrow \dfrac{x-8}{12x}<0 \\\Leftrightarrow \left[ \begin{array}{l}\begin{cases}x-8<0\\\\12x >0\end{cases}\\\begin{cases} x-8>0\\\\12x<0\end{cases}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\begin{cases}x<8\\\\x>0\end{cases}\\\begin{cases} x>8\\\\x<0\end{cases}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}0<x<8\\\\x\in \emptyset \end{array} \right.\\b.\\\dfrac{x-2}{3x} >\dfrac{1}{6}\\(x\ne0)\\ \Leftrightarrow \dfrac{2(x-2)-x}{6x} >0\\\Leftrightarrow \dfrac{x-4}{6x} >0 \\\Leftrightarrow \left[ \begin{array}{l}\begin{cases}x-4>0\\\\6x>0\end{cases}\\\begin{cases} x-4<0\\\\6x<0\end{cases}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\begin{cases}x>4\\\\x>0\end{cases}\\\begin{cases} x<4\\\\x<0\end{cases}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x<4\\\\x>0\end{array} \right.\)

Bài 3:

\(a.\\|-5x -2 |=\dfrac{1}{5}\\\Leftrightarrow \left[ \begin{array}{l}-5x -2 =\dfrac{1}{5}\\\\-5x -2 =-\dfrac{1}{5}\end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}-5x =\dfrac{11}{5}\\\\-5x =\dfrac{9}{5}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x = -\dfrac{11}{25}\\\\x =-\dfrac{9}{25}\end{array} \right.\\b.\\|3x+2|=-3\)

Khẳng định này là sai với mọi `x` vì hàm trị tuyệt đối luôn luôn dương hoặc bằng `0`

\(\Leftrightarrow x\in \emptyset\)

\(c.\\|-4x + 3|=2x +1\\\Leftrightarrow |-4x + 3 | -2x =1 \\\Leftrightarrow \left[ \begin{array}{l}-4x + 3 -2x =1\\\\-(-4x + 3)-2x =1\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}-6x =-2\\\\2x=4\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{1}{3}\\\\x=2\end{array} \right.\\d.\\|3-5x|=-2x -5\\|3-5x|+2x=-5\\\Leftrightarrow \left[ \begin{array}{l}3-5x + 2x = -5,\quad 3-5x \ge 0\\\\-(3-5x)+2x =-5,\quad 3-5x <0\end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}-3x =-8,\quad 3-5x \ge 0 \\\\7x =-2, \quad 3-5x<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x= \dfrac{8}{3}, \quad x\le \dfrac{3}{5}\\\\ x=-\dfrac {2}{7},\quad x >\dfrac{3}{5}\end{array} \right.\\\Rightarrow x\in \emptyset\)