Question

\(4-\left(4x-\dfrac{3}{2}\right)^2=0\) 

Xét 2 th <33

Answer 1

`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}`

Answer 2

=> \(\left(4x-\dfrac{3}{2}\right)^2=4\)

=> \(\left[{}\begin{matrix}4x-\dfrac{3}{2}=2\\4x-\dfrac{3}{2}=-2\end{matrix}\right.=>\left[{}\begin{matrix}4x=\dfrac{7}{2}\\4x=\dfrac{-1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)

Answer 3

 

 

Answer 4

\(4-\left(4x-\dfrac{3}{2}\right)^2=0\)

\(4x-\dfrac{3}{2}=2\)
4−(4x−32)2 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: \(4x-\dfrac{3}{2}=-2\)

         \(4x=-2+\dfrac{3}{2}\)

         \(4x=\dfrac{-1}{2}\)

         \(x=\dfrac{-1}{2}\div4\)

         \(x=\dfrac{-1}{8}\)