\(2x^2-5x+7=0\)
\(\Leftrightarrow2.\left[\left(x^2-2.x.\frac{5}{4}+\frac{25}{16}\right)+\frac{31}{16}\right]=0\)
\(\Leftrightarrow2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}=0\)
\(\Leftrightarrow2\left(x-\frac{5}{4}\right)^2=-\frac{31}{8}\) ( vô lí )
Do \(2\left(x-\frac{5}{4}\right)^2\ge0\forall x,-\frac{31}{8}< 0\)
Vậy : \(x\in\varnothing\)