Question

Chứng minh rằng :

x^2 + 3 - x > 0 vs mọi x € R

Answer 1

đặt A = x^2 + 3 - x

\(A=x^2+3-x\\ =x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}-\dfrac{1}{4}+3\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

vậy Min A = \(\dfrac{11}{4}\) khi x = \(\dfrac{1}{2}\)

vậy A > 0 với mọi x thuộc R

Answer 2

Ta có: \(x^2+3-x\)

\(=x^2-x+3\)

\(=x^2-2\cdot x\cdot1,5+2,25+0,75\)

\(=\left(x-1,5\right)^2+0,75\)

Vì \(\left(x-1,5\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1,5\right)^2+0,75\ge0,75>0\forall x\)

Vậy \(x^2+3-x>0\forall x\in R\)

Answer 3