a/ \(x^2-6x+10=x^2-2.x.3+3^2+1=\left(x-3\right)^2+1\)
Với mọi x ta có :
\(\left(x-3\right)^2\ge0\)
\(\Leftrightarrow\left(x-3\right)^2+1>0\)
\(\Leftrightarrow x^2-6x+10>0\)
b/ \(x^2-4x+7=x^2-2.x.2+2^2+3=\left(x-2\right)^2+3\)
Với mọi x ta có :
\(\left(x-2\right)^2\ge0\)
\(\Leftrightarrow\left(x-2\right)^2+3\ge3\)
\(\Leftrightarrow x^2-4x+7\ge3\left(đpcm\right)\)
c/ \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Với mọi x ta có :
\(\left(x+\dfrac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Leftrightarrow x^2+x+1>0\left(đpcm\right)\)
d/ \(x^2+y^2+4x-6y+15=\left(x^2+4x+2^2\right)+\left(y^2-6y+3^2\right)+2=\left(x+2\right)^2+\left(y-3\right)^2+2\)
Với mọi x,y ta có :
\(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2+2\ge0\)
\(\Leftrightarrow x^2+y^2+4x-6y+15>0\left(đpcm\right)\)
