a)
\(|x+1|=\left[ \begin{array}{l}x+1\\-x-1\end{array} \right.\)
⇒ TH1: $2x-(x+1)=\dfrac{-1}{2}$
$2x-x-1=\dfrac{-1}{2}$
$x=\dfrac{-1}{2}+1=\dfrac{1}{2}$ (thỏa mãn)
⇒ TH2: $2x-(-x-1)=\dfrac{-1}{2}$
$2x+x+1=\dfrac{-1}{2}$
$3x=\dfrac{-1}{2}-1=\dfrac{-3}{2}$
$x=\dfrac{-3}{2}.\dfrac{1}{3}=\dfrac{-1}{2}$ (không thỏa mãn)
b)
\(|x+15|=\left[ \begin{array}{l}x+15\\-x-15\end{array} \right.\)
⇒ TH1: $3x-(x+15)=\dfrac{5}{4}$
$3x-x-15=\dfrac{5}{4}$
$2x=\dfrac{5}{4}+15=\dfrac{65}{4}$
$x=\dfrac{65}{8}$ (thỏa mãn)
⇒ TH2: $3x-(-x-15)=\dfrac{5}{4}$
$3x+x+15=\dfrac{5}{4}$
$4x=\dfrac{-55}{4}$
$x=\dfrac{-55}{16}$ (không thỏa mãn)
