Question

(x+\(\dfrac{1}{3}\))\(^2\)=\(\dfrac{4}{9}\) Tìm x

Answer 1

`(x + 1/3)^2 = 4/9`

`=> (x + 1/3)^2 = (+-2/3)^2`

`=> x + 1/3 = 2/3` hoặc `x + 1/3 =-2/3`

`=> x = 2/3 - 1/3` hoặc `x = -2/3 - 1/3`

`=> x = 1/3` hoặc `x = -1`

Vậy `x \in {1/3;-1}`

Answer 2

\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{4}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{2}{3}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\\ \text{vậy }x\in\left\{\dfrac{1}{3};-1\right\}\)