Ta có:
\(9x^2-6x+1=0\)
\(\Leftrightarrow\left(3x-1\right)^2=0\)
\(\Leftrightarrow3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy để \(9x^2-6x+1=0\) thì \(x=\dfrac{1}{3}\)
\(9x^2-6x+1=0\)
\(\Leftrightarrow\left(3x\right)^2-2.3.x+1^2=0\)
\(\Leftrightarrow\left(3x-1\right)^2=0\)
\(\Rightarrow3x-1=0\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\dfrac{1}{3}\)