`4-(4x-3/2)^2=0`
`(4x-3/2)^2=4`
`@TH1:(4x-3/2)^2=2^2`
`=>4x-3/2=2`
`=>4x=7/2`
`=>x=7/8`
`@TH2:(4x-3/2)^2=(-2)^2`
`=>4x-3/2=-2`
`=>4x=-1/2`
`=>x=-1/8`
Vậy `x in {7/8;-1/8}`
=> \(\left(4x-\dfrac{3}{2}\right)^2=4\)
=> \(\left[{}\begin{matrix}4x-\dfrac{3}{2}=2\\4x-\dfrac{3}{2}=-2\end{matrix}\right.=>\left[{}\begin{matrix}4x=\dfrac{7}{2}\\4x=\dfrac{-1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
\(4-\left(4x-\dfrac{3}{2}\right)^2=0\)
hoặc \(4x-\dfrac{3}{2}=-2\)
TH1: \(4x-\dfrac{3}{2}=2\)
\(4x=2+\dfrac{3}{2}\)
\(4x=\dfrac{7}{2}\)
\(x=\dfrac{7}{2}\div4\)
\(x=\dfrac{7}{8}\)
TH2: \(4x-\dfrac{3}{2}=-2\)
\(4x=-2+\dfrac{3}{2}\)
\(4x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}\div4\)
\(x=\dfrac{-1}{8}\)
