\(A=\dfrac{x^2+x+1}{x^2+2x+1}=1-\dfrac{x}{\left(x+1\right)^2}\)
\(=1-\dfrac{x+1-1}{\left(x+1\right)^2}\)
\(=1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\)
Đặt 1/x+1=a
\(A=a^2-a+1=\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu = xảy ra khi a=1/2
=>1/x+1=1/2
=>x=1