`(x+1/2)^2 = 4/9`
`(x+1/2)^2 = (2/3)^2`
`@TH1:`
`x+1/2=2/3`
`x=2/3-1/2`
`x=4/6-3/6`
`x=1/6`
`@TH2:`
`x+1/2=-2/3`
`x=-2/3-1/2`
`x=-4/6-3/6`
`x=-7/6`
Vậy `x = {1/6;-7/6}`
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)
\(x+\dfrac{1}{2}=\dfrac{2}{3}\) hoặc \(x+\dfrac{1}{2}=\dfrac{-2}{3}\)
*) \(x+\dfrac{1}{2}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}-\dfrac{1}{2}\)
\(x=\dfrac{1}{6}\)
*) \(x+\dfrac{1}{2}=\dfrac{-2}{3}\)
\(x=\dfrac{-2}{3}-\dfrac{1}{2}\)
\(x=\dfrac{-7}{6}\)
Vậy \(x=\dfrac{-7}{6}\); \(x=\dfrac{1}{6}\)
\(TH1\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\\ x+\dfrac{1}{2}=\dfrac{2}{3}\\ x=\dfrac{2}{3}-\dfrac{1}{2}\\ x=\dfrac{2.2-1.3}{6}\\ x=\dfrac{1}{6}\\ TH2\left(x+\dfrac{1}{2}\right)^2=\left(-\dfrac{2}{3}\right)^2\\ x+\dfrac{1}{2}=-\dfrac{2}{3}\\ x=-\dfrac{2}{3}-\dfrac{1}{2}\\ x=-\dfrac{7}{6}\)
`(x+1/2)^2 = 4/9`
`=> TH1 : x+ 1/2 = sqrt{4/9} = 2/3 `
` = > x = 2/3 -1/2 = 4/6 - 3/6 =1/6`
`TH2 : x+1/2 = -sqrt{4/9} = -2/3`
`=> x = -2/3 -1/2 = -4/6 -3/6 =-7/6`
Vậy `x in {1/6;-7/6}`
(x + `1/2`)2 = `4/9`
x + `1/2` = `2/3`
x + `1/2` = `-2/3`
=> x = `2/3` - `1/2`
x = `1/6`
=> x = `-2/3` - `1/2`
x = `-7/6`
Vậy: x = `1/6` hoặc `-7/6`.
(x+1/2)²=(2/3)²
=>\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{3}\\x+\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}-\dfrac{1}{2}\\x=-\dfrac{2}{3}-\dfrac{1}{2}\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=\dfrac{4}{6}-\dfrac{3}{6}\\x=-\dfrac{4}{6}-\dfrac{3}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)Vậy x thuộc {1/6;-7/6}
