`(2/5-3x)^2=9/25`
`(2/5-3x)^2=(3/5)^2` hoặc `(2/5-3x)^2=(-3/5)^2`
`@TH1:2/5-3x=3/5=>3x=2/5-3/5=-1/5`
`=>x=-1/5:3=-1/15`
`@TH2:2/5-3x=-3/5=>3x=2/5-(-3/5)=1`
`=>x=1/3`
=>3x-2/5=3/5 hoặc 3x-2/5=-3/5
=>3x=1 hoặc 3x=-1/5
=>x=-1/15 hoặc x=1/3
\(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}=\left(\pm\dfrac{3}{5}\right)^2\\ =>\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=\dfrac{2}{5}-\dfrac{3}{5}=-\dfrac{1}{5}\\3x=\dfrac{2}{5}-\left(-\dfrac{3}{5}\right)=1\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{3}{5}\right)^2\) hoặc \(\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{-3}{5}\right)^2\)
\(\Rightarrow\dfrac{2}{5}-3x=\dfrac{3}{5}\) hoặc \(\dfrac{2}{5}-3x=\dfrac{-3}{5}\)
*) \(\dfrac{2}{5}-3x=\dfrac{3}{5}\)
\(3x=\dfrac{2}{5}-\dfrac{3}{5}\)
\(3x=-\dfrac{1}{5}\)
\(x=-\dfrac{1}{5}:3\)
\(x=-\dfrac{1}{15}\)
*) \(\dfrac{2}{5}-3x=-\dfrac{3}{5}\)
\(3x=\dfrac{2}{5}+\dfrac{3}{5}\)
\(3x=1\)
\(x=\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{15}\); \(x=\dfrac{1}{3}\)
