tìm x ,bt : (x-⅓)² = 16/9

conan nek (。・ω・。)

$\left(x-\frac{1}{3}\right)^2=\frac{16}{9}$

$\left(x-\frac{1}{3}\right)^2=\left(\frac{4}{3}\right)^2$

$\left[\begin{matrix}x-\frac{1}{3}=\frac{4}{3} \\ x-\frac{1}{3}=-\frac{4}{3}\end{matrix}\right.$

$\left[\begin{matrix}x=\frac{5}{3} \\ x=-1\end{matrix}\right.$

Vậy $x\in\left\lbrace-1;\frac{5}{3}\right\rbrace$.