\(5.\left(0,6-x\right)^2=\dfrac{125}{9}\Leftrightarrow\left(0,6-x\right)^2=\dfrac{125}{9.5}=\dfrac{25}{9}\)
\(\Leftrightarrow\left\{{}\begin{matrix}0,6-x=\dfrac{5}{3}\\0,6-x=\dfrac{-5}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-16}{15}\\x=\dfrac{34}{15}\end{matrix}\right.\)
vậy \(x=\dfrac{-16}{15};x=\dfrac{34}{15}\)
\(5\left(0.6-x\right)^2=\dfrac{125}{9}\\ \Leftrightarrow\left(0.6-x\right)^2=\dfrac{125}{9\cdot5}\\ \Leftrightarrow\left(0.6-x\right)^2=\dfrac{25}{9}\\ \Leftrightarrow\left(0.6-x\right)^2=\left(\dfrac{5}{9}\right)^2\text{ hoặc : }\\ \left(0.6-x\right)^2=\left(-\dfrac{5}{9}\right)^2\\ \Leftrightarrow0.6-x=\dfrac{5}{9}\text{ hoặc : }\\ 0.6-x=-\dfrac{5}{9}\\ \Leftrightarrow\left[{}\begin{matrix}0.6-x=\dfrac{5}{9}\\0.6-x=-\dfrac{5}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{45}\\x=-\dfrac{52}{45}\end{matrix}\right.\)
Vậy \(x=-\dfrac{2}{45}\) hoặc \(x=-\dfrac{52}{45}\)
