\(x^2-3x+4=0\)
\(\Leftrightarrow\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{7}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}=0\) (vô lý)
Vì: \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\in R\) \(\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\forall x\in R\)
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