|x+2|−6x=1
-> |x+2|=6x+1(1)
TH1: Nếu x+2 \(\ge\)0 \(\Leftrightarrow\)x\(\ge\)-2 \(\Rightarrow\)|x+2|=x+2
Thay vào (1) ta có:
x+2 = 6x + 1
\(\Leftrightarrow\)x-6x=1-2
\(\Leftrightarrow\)-5x=-1
\(\Leftrightarrow\)x= \(\dfrac{1}{5}\)(TMĐK)
TH2: Nếu x+2\(\le\)0 \(\Leftrightarrow\)x < -2 \(\Rightarrow\)|x+2| = -(x+2)
Thay vào (1) ta có:
-x-2=6x+1
\(\Leftrightarrow\)-x-6x=1+2
\(\Leftrightarrow\)-7x=3
\(\Leftrightarrow\)x=\(-\dfrac{3}{7}\)(KTMĐK)
Vậy S={\(\dfrac{1}{5}\)}
