Lời giải:
\(-3x^2+8x-2=0\)
\(\Leftrightarrow 3x^2-8x+2=0\)
\(\Leftrightarrow 3(x^2-\frac{8}{3}x+\frac{8^2}{6^2})=\frac{10}{3}\)
\(\Leftrightarrow 3(x-\frac{8}{6})^2=\frac{10}{3}\)
\(\Leftrightarrow (x-\frac{4}{3})^2=\frac{10}{9}\Rightarrow \left[\begin{matrix} x-\frac{4}{3}=\frac{\sqrt{10}}{3}\\ x-\frac{4}{3}=\frac{-\sqrt{10}}{3}\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{4+\sqrt{10}}{3}\\ x=\frac{4-\sqrt{10}}{3}\end{matrix}\right.\)
\(-3x^2+8x-2=0\)
\(\Leftrightarrow-3\left(x^2-\dfrac{8}{3}x+\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow-3\left(x^2-2x.\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{10}{9}\right)=0\)
\(\Leftrightarrow-3\left[\left(x-\dfrac{4}{3}\right)^2-\dfrac{10}{9}\right]=0\)
\(\Leftrightarrow\left(x-\dfrac{4}{3}\right)^2-\dfrac{10}{9}=0\)
\(\Leftrightarrow\left(x-\dfrac{4}{3}\right)^2=\dfrac{10}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{4}{3}=\dfrac{\sqrt{10}}{3}\\x-\dfrac{4}{3}=\dfrac{-\sqrt{10}}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}+4}{3}\\x=\dfrac{4-\sqrt{10}}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}+4}{3}\\x=\dfrac{4-\sqrt{10}}{3}\end{matrix}\right.\)
