a: \(A=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
b: \(B=-x^2+4x-17\)
\(=-\left(x^2-4x+17\right)\)
\(=-\left(x^2-4x+4+13\right)\)
\(=-\left(x-2\right)^2-13< 0\forall x\)
a) \(A=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
b) \(4x-17-x^2=-\left(x^2-4x+4\right)-13=-\left(x-2\right)^2-13\le-13< 0\)
a) A = \(x^2-x+1\)
= \(x^2\) - 2.\(x\).\(\dfrac{1}{2}\) + \(\left(\dfrac{1}{2}\right)^2\) + \(\dfrac{3}{4}\)
= \(\left(x-\dfrac{1}{2}\right)^2\) + \(\dfrac{3}{4}\)
Với mọi \(x\) ta có:
\(\left(x-\dfrac{1}{2}\right)^2\) ≥ 0
➩\(\left(x-\dfrac{1}{2}\right)^2\) + \(\dfrac{3}{4}\) > \(\dfrac{3}{4}\)
➩\(\left(x-\dfrac{1}{2}\right)^2\) + \(\dfrac{3}{4}\) > 0
➩\(x^2-x+1\) > 0
➩ A > 0
Vậy biểu thức A = \(x^2-x+1\) luôn dương với mọi \(x\)
