\(P=x^2-2x+2\)
\(P=x^2-2x+1+1\)
\(P=\left(x-1\right)^2+1\)
Ta thấy \(\left(x-1\right)^2\ge0\)nên \(\left(x-1\right)^2+1>0\)
Ta có:
\(P=x^2-2x+2\)
\(=\left(x^2-2x+1\right)+1\)
\(=\left(x-1\right)^2+1\)
Vì \(\left(x-1\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+1>0\forall x\)