Question

chứng minh rằng x²-5x+7>0 với mọi x

Answer 1

\(x^2-5x+7\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{25}{4}+7\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)

Ta thấy: \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

hay \(x^2-5x+7>0\forall x\).

Vậy ...

#\(Toru\)