x2+2x+11=(x2+2x+1)+10=(x+1)2+10
Do (x+1)2>0=>(x+1)2+10>0
=>(x+1)2+10 luôn dương
=>x2+2x+11 luôn dương(đpcm)
x^2+2x+11
=x2+2x+1+10
=(x+1)2+10 >0 với mọi x ( vì (x+1)2\(>\)0)
vậy x^2+2x+11 luôn dương
\(=\left(x+1\right)^2+10\ge10\)
Vậy bài toán trên luôn dương với mọi x
TA CÓ:\(x^2+2x+11=x^2+2.1.x+1^2+10=\left(x^2+2.1.x+1^2\right)+10=\left(x+1\right)^2+10\)
do:\(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+10\ge10>0\left(đpcm\right)\)
\(x^2+2x+11\)
\(=x^2+2.1.x+1+10\)
\(=\left(x^2+2.1.x+1^2\right)+10\)
\(=\left(x+1\right)^2+10\)
do \(\left(x+1\right)^2\ge0,\)\(10>0\)
\(\Rightarrow\left(x+1\right)^2+10>0\)
\(\Rightarrow\)\(x^2+2x+11\)luôn dương
