`|x-1/2| =3/4`
`TH1: x-1/2 = 3/4`
`=>x= 3/4 + 1/2`
`=>x=5/4`
`TH2:x- 1/2 = -3/4`
`=>x= -3/4 + 1/2`
`=>x= -1/4`
Vậy `x in {-1/4 ; 5/4}`
\(\left|x-\dfrac{1}{2}\right|=\dfrac{3}{4}< =>\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{4}\\x-\dfrac{1}{2}=-\dfrac{3}{4}\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{3}{4}+\dfrac{1}{2}\\x=-\dfrac{3}{4}+\dfrac{1}{2}\end{matrix}\right.\)
\(< =>\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\( x - \dfrac{1}{2} | = \dfrac{3}{4} \)
\(TH1\)
\(x-\dfrac{1}{2}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}+\dfrac{1}{2}\)
\(x=\dfrac{3}{4}+\dfrac{2}{4}\)
\(x=\dfrac{5}(4}\)
\(TH2\)
\(x-\dfrac{1}{2}=\dfrac{-3}{4}\)
`=> x=-3/4+1/2`
`=> x=-3/4+2/4`
`=> x=-1/4`
Vậy `x∈{5/4; -1/4}`
