Ta có: \(x^2-x+\frac{7}{4}\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{2}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{2}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}>0\forall x\)
hay \(x^2-x+\frac{7}{4}>0\forall x\)
Vậy: Đa thức \(x^2-x+\frac{7}{4}\) vô nghiệm(đpcm)