Ta có :
\(A=\frac{x^2+x+1}{\left(x+1\right)^2}\)
\(A=\frac{x^2+2x+1-x-1+1}{x^2+2x+1}\)
\(A=\frac{x^2+2x+1}{\left(x+1\right)^2}+\frac{-x-1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\)
\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)^2}-\frac{x+1}{\left(x+1\right)^2}+\frac{1^2}{\left(x+1\right)^2}\)
\(A=1-\frac{1}{x+1}+\left(\frac{1}{x+1}\right)^2\)
Đặt \(a=\frac{1}{x+1}\) ta có :
\(A=1-a+a^2\)
\(A=a^2-a+1\)
\(A=\left(a^2-a+\frac{1}{4}\right)+\frac{3}{4}\)
\(A=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra khi và chỉ khi \(\left(a-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\)\(a-\frac{1}{2}=0\)
\(\Leftrightarrow\)\(a=\frac{1}{2}\)
Do đó :
\(a=\frac{1}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{2}=\frac{1}{x+1}\)
\(\Leftrightarrow\)\(x+1=2\)
\(\Leftrightarrow\)\(x=1\)
Vậy GTNN của \(A\) là \(\frac{3}{4}\) khi \(x=1\)
Chúc bạn học tốt ~
