Phương trình $\Large \dfrac{1}{3}-\left| \dfrac{5}{4}-2x \right|=\dfra

Phương trình $\Large \dfrac{1}{3}-\left| \dfrac{5}{4}-2x \right|=\dfrac{1}{4}$ có nghiệm là:

• A. $x=\dfrac{7}{12};x=\dfrac{1}{12}$
• B. $x=\dfrac{8}{12};x=\dfrac{5}{12}$
• C. $x=\dfrac{7}{12};x=\dfrac{2}{3}$
• D. $x=\dfrac{1}{12};x=\dfrac{5}{12}$

$\Large \begin{align}  & \dfrac{1}{3}-\left| \dfrac{5}{4}-2x \right|=\dfrac{1}{4} \\  & \Leftrightarrow \left| \dfrac{5}{4}-2x \right|=\dfrac{1}{3}-\dfrac{1}{4} \\  & \Leftrightarrow \left| \dfrac{5}{4}-2x \right|=\dfrac{1}{12} (*) \\ \end{align}$

TH1: $\Large \dfrac{5}{4}-2x\ge 0\Leftrightarrow x\le \dfrac{5}{8}\Rightarrow \left| \dfrac{5}{4}-2x \right|=\dfrac{5}{4}-2x$

$\Large \begin{align} \Rightarrow \left( * \right) & \Leftrightarrow \dfrac{5}{4}-2x=\dfrac{1}{12} \\  & \Leftrightarrow 2x=\dfrac{7}{6} \\  & \Leftrightarrow x=\dfrac{7}{12}(tm) \\ \end{align}$

TH2: $\Large \dfrac{5}{4}-2x<0\Leftrightarrow x>\dfrac{5}{8}\Rightarrow \left| \dfrac{5}{4}-2x \right|=-\dfrac{5}{4}+2x$

$\Large \begin{align} \Rightarrow \left( * \right)&\Leftrightarrow -\dfrac{5}{4}+2x=\dfrac{1}{12} \\ & \Leftrightarrow 2x=\dfrac{4}{3} \\  & \Leftrightarrow x=\dfrac{2}{3}(tm) \\ \end{align}$

Vậy phương trình có hai nghiệm $\Large \dfrac{7}{12}$ và $\Large \dfrac{2}{3}$