a, => x - 1,5 = \(\pm\) 2
\(\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
b, \(|x+\dfrac{3}{4}|=0+\dfrac{1}{2}\)
\(|x+\dfrac{3}{4}|=\dfrac{1}{2}\)
=> x + \(\dfrac{3}{4}=\pm\dfrac{1}{2}\)
\(\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{2}\\x+\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-5}{4}\end{matrix}\right.\)
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