\(x^2-x+3\)\(=x^2-x+\dfrac{1}{4}+\dfrac{11}{4}\)\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}>0\)(luôn đúng)Câu trả lời: \(x^2-x+3\)\(=x^2-x+\dfrac{1}{4}+\dfrac{11}{4}\)\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}>0\)(luôn đúng)