a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=4\\\left(4x\right)^2-\left(x-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=4\\\left(3x+4\right)\left(5x-4\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
b: =>2x+3=x-1 hoặc 2x+3=1-x
=>x=-4 hoặc 3x=-2
=>x=-4 hoặc x=-2/3
`a)`\(\left|4x\right|=x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=x-4;x\ge0\\4x=4-x;x< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\left(ktm\right)\\x=\dfrac{4}{5}\left(ktm\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
`b)`\(\left|x-1\right|=\left|2x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x+3\\1-x=2x+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-4;-\dfrac{2}{3}\right\}\)
