Đáp án:
$4-\left | \dfrac{2}{3}x+1 \right |=6\Leftrightarrow x\in \varnothing$
$\left ( \dfrac{1}{2}x+1 \right ).3-\left ( \dfrac{1}{2}x+1 \right ):4.x=0\Leftrightarrow x=-2$
$\left | \dfrac{1}{x}-\dfrac{1}{3} \right |-\dfrac{1}{2}=\dfrac{1}{3}\Leftrightarrow x\in \left \{ \dfrac{6}{7};-2 \right \}$
Giải thích các bước giải:
$4-\left | \dfrac{2}{3}x+1 \right |=6$
$\Leftrightarrow \left | \dfrac{2}{3}x+1 \right |=-2$
$\Leftrightarrow x\in \varnothing$ (vì $\left | \dfrac{2}{3}x+1 \right |\geq 0\forall x$)
$\left ( \dfrac{1}{2}x+1 \right ).3-\left ( \dfrac{1}{2}x+1 \right ):4.x=0$
$\Leftrightarrow \left ( \dfrac{1}{2}x+1 \right )\left ( 3-\dfrac{x}{4} \right )=0$
$\Leftrightarrow \dfrac{1}{2}x+1=0$
$\Leftrightarrow \dfrac{1}{2}x=-1$
$\Leftrightarrow x=-2$
$\left | \dfrac{1}{x}-\dfrac{1}{3} \right |-\dfrac{1}{2}=\dfrac{1}{3}$
$\Leftrightarrow \left | \dfrac{1}{x}-\dfrac{1}{3} \right |=\dfrac{5}{6}$
$\Leftrightarrow \left[ \begin{array}{l}\dfrac{1}{x}-\dfrac{1}{3}=\dfrac{5}{6}\\\dfrac{1}{x}-\dfrac{1}{3}=-\dfrac{5}{6}\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}\dfrac{1}{x}=\dfrac{7}{6}\\\dfrac{1}{x}=-\dfrac{1}{2}\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{6}{7}\\x=-2\end{array} \right.$
