\(M=3x^2+6x+9\)
\(M=3\left(x^2+2x+3\right)\)
\(M=3\left(x^2+2x+1+2\right)\)
\(M=3\left[\left(x+1\right)^2+2\right]\)
\(M=3\left(x+1\right)^2+6\)
\(\left(x+1\right)^2\ge0\)
\(\Rightarrow3\left(x+1\right)^2\ge0\)
\(\Rightarrow3\left(x+1\right)^2+6\ge6\)
Vậy biểu thức M luôn luôn dương \(\forall x\)
\(M=3x^2+6x+9=3x^2+6x+3+6\)
\(=3\left(x^2+2x+1\right)+6\)\(=3\left(x+1\right)^2+6\)
Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow3\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow M\ge6\forall x\)\(\Rightarrow\)M luôn dương ( đpcm )
Ta có : \(M=3x^2+6x+9\)
\(=\left(3x^2+6x+3\right)+6\)
\(=3\left(x^2+2x+1\right)+6=3\left(x+1\right)^2+6\)
Ta thấy \(\left(x+1\right)^2\ge0\forall x\)
=> \(3\left(x+1\right)^2 \ge0\forall x\)
=> \(3\left(x+1\right)^2+6 >0\forall x\)
hay M luôn dương (đpcm)
\(M=3x^2+6x+9\)
\(=\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.\frac{3}{\sqrt{3}}+3+6\)
\(=\left(\sqrt{3}x+\frac{\sqrt{3}}{3}\right)^2+6>0\forall x\)
Vậy \(M=3x^2+6x+9\)luôn dương