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