\(\left|x-7\right|=2x+3\)
\(x-7=2x+3\)
\(x-2x=3+7\)
\(-x=10\)
\(x=-10\)
\(\left|x-7\right|=2x+3\)
\(\Leftrightarrow x-7=2x+3\)
\(\Leftrightarrow-x-7=3\)
\(\Leftrightarrow-x=10\)
\(\Leftrightarrow x=-10\)
\(\left|x-7\right|=2x+3\)
\(\Leftrightarrow\)\(\left|x-7\right|=-\left(2x+3\right)\)
\(\Leftrightarrow\)\(\left|x-7\right|=-2x-3\)
\(\Leftrightarrow\)\(3x-7=-3\)
\(\Leftrightarrow\)\(3x=\left(-3\right)+7\)
\(\Leftrightarrow\)\(3x=4\)
\(\Leftrightarrow\) \(x=\dfrac{4}{3}\)
Vậy \(x=\dfrac{4}{3}\)