\(A=\dfrac{4\left(x^2+2x+3-3\right)+18}{x^2+2x+3}=\dfrac{4\left(x^2+2x+3\right)+6}{x^2+2x+3}=4+\dfrac{6}{\left(x+1\right)^2+2}\)
Ta có \(\left(x+1\right)^2+2\ge2\Rightarrow\dfrac{6}{\left(x+1\right)^2+2}\le3\Leftrightarrow4+\dfrac{6}{\left(x+1\right)^2+2}\le7\)
Dấu ''='' xảy ra khi x = -1