\(|4x-3|-|x+7|=0\)
\(\Leftrightarrow|4x-3|=|x+7|\)
\(\Leftrightarrow\orbr{\begin{cases}4x-3=x+7\\4x-3=-x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=10\\5x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=\frac{-4}{5}\end{cases}}\)
Vậy \(x\in\left\{\frac{10}{3};\frac{-4}{5}\right\}\)
\(\left|4x-3\right|-\left|x+7\right|=0\)
\(\Leftrightarrow\left|4x-3\right|=\left|x+7\right|\)(1)
* Nếu \(x\ge-7\)thì \(x+7\ge0\Rightarrow\left|x+7\right|=x+7\)
\(\Rightarrow\left(1\right)\Leftrightarrow x+7=\left|4x-3\right|\)(2)
+) Nếu \(x\ge\frac{4}{3}\)thì \(\left(2\right)\Leftrightarrow x+7=4x-3\Leftrightarrow x=\frac{10}{3}\left(TM\right)\)
+) Nếu \(x< \frac{4}{3}\)thì \(\left(2\right)\Leftrightarrow x+7=3-4x\Leftrightarrow x=\frac{-4}{5}\left(TM\right)\)
* Nếu \(x< -7\)thì \(x+7< 0\Rightarrow\left|x+7\right|=-x-7\)
\(\Rightarrow\left(1\right)\Leftrightarrow-x-7=\left|4x-3\right|\)(3)
+) Nếu \(x\ge\frac{4}{3}\)thì \(\left(3\right)\Leftrightarrow-x-7=4x-3\Leftrightarrow x=\frac{-4}{5}\left(TM\right)\)
+) Nếu \(x< \frac{4}{3}\)thì \(\left(3\right)\Leftrightarrow-x-7=3-4x\Leftrightarrow x=\frac{3}{10}\left(TM\right)\)
Vậy \(x\in\left\{\frac{-4}{5};\frac{3}{10}\right\}\)