Question

giúp mk vs

Answer 1

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

\(\Leftrightarrow\left(x-1\dfrac{1}{3}\right)^2=\left(\dfrac{4}{3}\right)^2\)

\(\Rightarrow x-\dfrac{4}{3}=\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{4}{3}+\dfrac{4}{3}=\dfrac{8}{3}\)
Học tốt

Answer 2

Ta có:

\(\dfrac{16}{9}=\left(\dfrac{4}{3}\right)^2=\left(-\dfrac{4}{3}\right)^2\)

Theo đề đầu bài:

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

\(\left(x-\dfrac{4}{3}\right)^2=\dfrac{16}{9}\)

Vậy \(x-\dfrac{4}{3}=\dfrac{4}{3}\) hoặc \(x-\dfrac{4}{3}=-\dfrac{4}{3}\)

Nếu \(x-\dfrac{4}{3}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{8}{3}\)

Nếu \(x-\dfrac{4}{3}=-\dfrac{4}{3}\Leftrightarrow x=0\)

Vậy \(x=0,x=\dfrac{8}{3}\)