\(\left(x+\dfrac{1}{2}\right)^2-\dfrac{49}{36}=0\)
\(\left(x+\dfrac{1}{2}\right)^2\) \(=0+\dfrac{49}{36}=\perp\left(\dfrac{7}{6}\right)^2\)
\(\text{Vậy }x+\dfrac{1}{2}=\dfrac{7}{6}\)
\(x\) \(=\dfrac{7}{6}+\left(\dfrac{-1}{2}\right)=\dfrac{2}{3}\)
\(\text{hoặc }x+\dfrac{1}{2}=\left(\dfrac{-7}{6}\right)\)
\(x\) \(=\left(\dfrac{-7}{6}\right)+\left(\dfrac{-1}{2}\right)=\dfrac{-5}{3}\)
\(\Rightarrow x\in\left\{\dfrac{2}{3};\left(\dfrac{-5}{3}\right)\right\}\)
(x+1/2)^2-49/36=0
(x+1/2)^2=0+49/36
(x+1/2)^2=49/36
(x+1/2)^2=(+-7/6)^2
x+1/2=7/6 hoặc x+1/2=(-7/6)
=>x=2/3 hoặc x=-5/3