\(|x+2|-6x=1\)
\(\Leftrightarrow x+2-6x=1\)
\(\Leftrightarrow x-6x=1-2\)
\(\Leftrightarrow-5x=-1\)
\(\Leftrightarrow x=\dfrac{-1}{-5}\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
Vậy \(x=\dfrac{1}{5}\)
\(\left|x+2\right|-6x=1\)
\(th1:x+2\ge0\Leftrightarrow x\ge-2\)
\(\Rightarrow\left|x+2\right|-6x=1\Leftrightarrow x+2-6x=1\Leftrightarrow5x=1\Leftrightarrow x=\dfrac{1}{5}\left(tmđk\right)\)
\(th2:x+2< 0\Leftrightarrow x< -2\)
\(\Rightarrow\left|x+2\right|-6x=1\Leftrightarrow-x-2-6x=1\Leftrightarrow-7x=3\Leftrightarrow x=\dfrac{-3}{7}\left(loại\right)\)
vậy \(x=\dfrac{1}{5}\)
\(\left|x+2\right|-6x=1\)
\(\Rightarrow\left|x+2\right|=6x+1\)
\(\Rightarrow\left[{}\begin{matrix}x+2=6x+1\forall x\ge-2\\-x-2=6x+1\forall x< -2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6x-1\Rightarrow5x=1\Rightarrow x=\dfrac{1}{5}\left(TM\right)\\-x=6x+3\Rightarrow-7x=3\Rightarrow x=-\dfrac{7}{3}\left(TM\right)\end{matrix}\right.\)
Ta có : | x + 2 | - 6x = 1
=> | x + 2 | = 6x + 1
=> \(\left[{}\begin{matrix}x+2=6x+1\ge-2\\-x-2=6x+1< -2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=6x-1=>5x=1=>x=\dfrac{1}{5}\left(TM\right)\\-x=6x+3=>-7x=3=>x=\dfrac{-7}{3}\left(TM\right)\end{matrix}\right.\)
Vậy ..............
