a,A=\(\sqrt{x^2+2x+5}=\sqrt{x^2+2x+1+4}=\sqrt{\left(x+1\right)^2+4}\)
ta có \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+4\ge4\)
\(\Rightarrow\)\(\sqrt{\left(x+1\right)^2+4}\ge2\) hay A\(\ge\)2
dấu ''='' xảy ra \(\Leftrightarrow\)x+1=0->x=1
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