Tìm x a,|2x-1/3|-|x+1/4|=0 b,5-|x-1|=2x

Đáp án:$\begin{array}{l}
a)x = \dfrac{7}{{12}};x = \dfrac{1}{{36}}\\
b)x = 2
\end{array}$

Giải thích các bước giải:

$\begin{array}{l}
a)\left| {2x - \dfrac{1}{3}} \right| - \left| {x + \dfrac{1}{4}} \right| = 0\\
 \Leftrightarrow \left| {2x - \dfrac{1}{3}} \right| = \left| {x + \dfrac{1}{4}} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{1}{3} = x + \dfrac{1}{4}\\
2x - \dfrac{1}{3} =  - x - \dfrac{1}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - x = \dfrac{1}{4} + \dfrac{1}{3}\\
2x + x =  - \dfrac{1}{4} + \dfrac{1}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{{12}}\\
3x = \dfrac{1}{{12}}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{{12}}\\
x = \dfrac{1}{{36}}
\end{array} \right.\\
Vậy\,x = \dfrac{7}{{12}};x = \dfrac{1}{{36}}\\
b)5 - \left| {x - 1} \right| = 2x\\
 \Leftrightarrow \left| {x - 1} \right| = 5 - 2x\left( {dk:x \le \dfrac{5}{2}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 5 - 2x\\
x - 1 = 2x - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2x = 5 + 1\\
2x - x =  - 1 + 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = 6\\
x = 4\left( {ktm} \right)
\end{array} \right.\\
 \Leftrightarrow x = 2\\
Vậy\,x = 2
\end{array}$