a) \(9x^2-6x+3=0\)
\(\Leftrightarrow\left(3x\right)^2-2.3x.1+1^2+2=0\)
\(\Leftrightarrow\left(3x-1\right)^2=-2\) ( vô lí )
b) \(x^2-7x+12=0\)
\(\Leftrightarrow x^2-2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2=\frac{1}{4}=\left(-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{2}=\frac{1}{2}\\x-\frac{7}{2}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}\)
Vậy : \(x\in\left\{3,4\right\}\)
c) \(x^2-8x+6=0\)
\(\Leftrightarrow x^2-2.x.4+4^2-10=0\)
\(\Leftrightarrow\left(x-4\right)^2=10\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=\sqrt{10}\\x-4=-\sqrt{10}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{10}+4\\x=-\sqrt{10}+4\end{cases}}\)
