Ta có :
\(Q=x^2-4x+7\)
\(=\left(x^2-4x+4\right)+3\)
\(=\left(x-2\right)^2+3\)
Với mọi x ta có :
\(\left(x-2\right)^2\ge0\)
\(\Leftrightarrow\left(x-2\right)^2+3\ge3\)
\(\Leftrightarrow Q\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow x=2\)
Vậy...
Ta có: \(Q=x^2-4x+7\)
\(\Leftrightarrow Q=x^2-4x+4+3\)
\(\Leftrightarrow Q=\left(x^2-4x+4\right)+3\)
\(\Leftrightarrow Q=\left(x-2\right)^2+3\)
Lại có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+3\ge3\forall x\)
Dấu \("="\) xảy ra \(x=2\)
Vậy ....................
