`#3107.101107`
\(1-\left(x+\dfrac{1}{3}\right)^2=\dfrac{3}{4}\\ \left(x+\dfrac{1}{3}\right)^2=1-\dfrac{3}{4}\\ \left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}\\ \left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\)
TH1:
\(x+\dfrac{1}{3}=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{1}{3}\\ x=\dfrac{1}{6}\)
TH2:
\(x+\dfrac{1}{3}=-\dfrac{1}{2}\\ x=-\dfrac{1}{2}-\dfrac{1}{3}\\ x=-\dfrac{5}{6}\)
Vậy, \(\in\left\{\dfrac{1}{6};-\dfrac{5}{6}\right\}.\)
`1- (x + 1/3)^2 = 3/4`
`=> (x+ 1/3)^2 = 1 - 3/4`
`=> (x+ 1/3)^2 = 1/4`
`=> (x + 1/3)^2 = (+-1/2)^2`
`=> x + 1/3 = 1/2` hoặc `x + 1/3 = -1/2`
`=> x= 1/2 - 1/3` hoặc `x = -1/2 - 1/3`
`=> x = 3/6 - 2/6` hoặc `x= -3/6 - 2/6`
`=> x=1/6` hoặc `x=-5/6`
Vậy: `x = 1/6; x = -5/6`
\(1-\left(x+\dfrac{1}{3}\right)^2=\dfrac{3}{4}\)
\(\left(x+\dfrac{1}{3}\right)^2=1-\dfrac{3}{4}\)
\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\left(x+\dfrac{1}{3}\right)^2=\left(\dfrac{1}{2}\right)^2\)
\(x+\dfrac{1}{3}=\dfrac{1}{2}\) hoặc \(x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{1}{3}\) hoặc \(x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) hoặc \(x=-\dfrac{5}{6}\)
