Ta có :
\(M=\frac{1}{x^2+4x+6}\)
\(=\frac{1}{\left(x^2+4x+4\right)+2}\)
\(=\frac{1}{\left(x+2\right)^2+2}\)
Với mọi x ta có :
\(\left(x+2\right)^2\ge0\)
\(\Leftrightarrow\left(x+2\right)^2+2\ge2\)
\(\Leftrightarrow\frac{1}{\left(x+2\right)^2+2}\le\frac{1}{2}\)
\(\Leftrightarrow M\le\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=-2\)
Vậy...