Đáp án:

`a)x∉∅`    `c)x∈{3/2;9/2}`      `c)x∈{2/5;6/7}`

Giải thích các bước giải:

`a)3x-|x-1|=0`

`→3x-x-1=0` 

`→2x=-1`

`→x=-1/2(x∈∅)`

`→`pt vô nghiệm

`b)|5-2x|-2/3x=1`

`→|-2x+5|=2/3x+1`

`→`\(\left[ \begin{array}{l}-2x+5=\dfrac{2}{3}x+1\\-2x+5=-(\dfrac{2}{3}x+1)\end{array} \right.\) 

`→`\(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{9}{2}\end{array} \right.\) 

`c)1/2x+|3x-2|=1`

`|3x-2|=1/2x+1`

`→`\(\left[ \begin{array}{l}3x-2=\dfrac{1}{2}x+1\\3x-2=-(\dfrac{1}{2}x+1)\end{array} \right.\)

`→`\(\left[ \begin{array}{l}x=\dfrac{2}{5}\\x=\dfrac{6}{7}\end{array} \right.\)

Đáp án+Giải thích các bước giải:

$a.\\3x -|x-1| =0\\\Leftrightarrow\left[ \begin{array}{l}3x -(x-1)=0, x-1\ge 0\\\\3x -[-(x-1)]=0, x-1<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}2x +1 =0, x\ge 1\\\\4x-1=0, x <1\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-\dfrac{1}{2}, x \ge 1\\\\x=\dfrac{1}{4}, x < 1\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x\in \emptyset \\\\x=\dfrac{1}{4}\end{array} \right.\\b.\\|5-2x| -\dfrac{2}{3} x =1\\\Leftrightarrow 3\times |5x -2x|-2x=3\\\Leftrightarrow \left[ \begin{array}{l}3(5-2x)-2x = 3, 5-2x \ge 0\\\\3[-(5-2x)]-2x =3, 5-2x <0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}15 - 6x - 2x =3, x \le \dfrac{5}{2}\\\\-15 + 6x - 2x =3, x > \dfrac{5}{2}\end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}-8x =-12x \le \dfrac{5}{2}\\\\x>\dfrac{5}{2}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{3}{2}, x\le \dfrac{5}{2}\\\\x=\dfrac{9}{2}, x>\dfrac{5}{2}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{3}{2}\\\\x=\dfrac{9}{2}\end{array} \right.\\c.\\\dfrac{1}{2}x + |3x-2|=1\\\Leftrightarrow x+2\times |3x-2|=2 \\\Leftrightarrow \left[ \begin{array}{l}x+2(3x-2)=2, 3x-2 \ge 0\\\\x+2 [-(3x-2)] =2, 3x-2<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}7x =6, x \ge \dfrac{2}{3}\\\\-5x =-2, x <\dfrac{2}{3}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{6}{7}\\x=\dfrac{2}{5}\end{array} \right. $