\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{2}=\dfrac{9}{2}\\ =>\left(x-\dfrac{3}{2}\right)^2=\dfrac{9}{2}-\dfrac{1}{2}\\ =>\left(x-\dfrac{3}{2}\right)^2=4\\ =>\left[{}\begin{matrix}x-\dfrac{3}{2}=2\\x-\dfrac{3}{2}=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
`(x-3/2)^2 = 4`
`<=>` \(\left[{}\begin{matrix}x-\dfrac{3}{2}=2\\x-\dfrac{3}{2}=-2\end{matrix}\right.\)
`<=>` \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{2}=\dfrac{9}{2}\)
\(\left(x-\dfrac{3}{2}\right)^2=\dfrac{9}{2}-\dfrac{1}{2}\\ \left(x-\dfrac{3}{2}\right)^2=4\\ \left(x-\dfrac{3}{2}\right)^2=2^2\)
\(x-\dfrac{3}{2}=2\)
\(x=2+\dfrac{3}{2}\)
\(x=\dfrac{7}{2}\)
Vậy \(x=\dfrac{7}{2}\)
\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{2}=\dfrac{9}{2}\)
\(\left(x-\dfrac{3}{2}\right)^2=\dfrac{9}{2}-\dfrac{1}{2}\)
\(\left(x-\dfrac{3}{2}\right)^2=\dfrac{8}{2}=4\)
\(\left(x-\dfrac{3}{2}\right)^2=\left(\pm2\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=2\\x-\dfrac{3}{2}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{7}{2};-\dfrac{1}{2}\right\}\).
