=>x+1/3=1/5 hoặc x+1/3=-1/5
=>x=-2/15 hoặc x=-8/15
`(x+1/3)^2 = 1/25`
`=>(x+1/3)^2 = (1/5)^2`
`=>` \(\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=-\dfrac{1}{5}\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=\dfrac{1}{5}-\dfrac{1}{3}\\x=-\dfrac{1}{5}-\dfrac{1}{3}\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=-\dfrac{8}{15}\end{matrix}\right.\)
Vậy....
\(\left(x+\dfrac{1}{3}\right)^2=\left(\dfrac{1}{5}\right)^2\)
\(x+\dfrac{1}{3}=\pm\dfrac{1}{5}\)
=> x + 1/3 = 1/5
x = 1/5 - 1/3 = -2/15
=> x + 1/3 = -1/5
x = -1/5 - 1/3 = -8/15
Vậy ...
( x + 1/3 )^2 = 1/25
TH1:
x + 1/3 = 1/5
x = -2/15
TH2:
x + 1/3 = -1/5
x = -8/15
\(\Rightarrow\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{1}{5}\right)^2\)
\(\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=-\dfrac{1}{5}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{5}-\dfrac{1}{3}\\x=-\dfrac{1}{5}-\dfrac{1}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-8}{15}\end{matrix}\right.\)
\(Vậy\) \(x=\dfrac{-2}{15}\) \(và\) \(x=\dfrac{-8}{15}\)