\(\left(x-1\dfrac{1}{3}\right)^2=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-1\dfrac{1}{3}\right)^2=\left(\dfrac{4}{3}\right)^2\)
\(\Rightarrow x-\dfrac{4}{3}=\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{4}{3}+\dfrac{4}{3}=\dfrac{8}{3}\)
Học tốt
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 \(x=0,x=\dfrac{8}{3}\)
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