ta có : \(x^3-\dfrac{1}{9}x=0\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\\x+\dfrac{1}{3}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\) vậy \(x=0;x=\dfrac{1}{3};x=\dfrac{-1}{3}\)
\(x^3-\dfrac{1}{9}x=0\\ \Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\\ \Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\\x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{1}{3}\) hoặc \(x=-\dfrac{1}{3}\)
a) x3 - \(\dfrac{1}{9}x\) = 0
x(x2 - \(\dfrac{1}{9}\)) = 0
x(x - \(\dfrac{1}{3}\))(x + \(\dfrac{1}{3}\)) = 0
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\\x+\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
\(x^3-\dfrac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{1}{3}=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\)
tik cho mik đi PLEASE, nhớ tik cho mik đấy, đừng có mà đọ suông xong bỏ đấy đấy
x^3- 1/9x=0
x (x^2-1/9)=0
TH1: x=0
TH2: x^2-1/9=0
x^2=1/9
x=1/3