Question

[giang ho dai ca]

Thank các bạn giải hộ

Answer 1

[Minh Triều]

\(\frac{x^2+x+1}{\left(x+1\right)^2}=\frac{x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}}{\left(x+1\right)^2}=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{\left(x+1\right)^2}\)

\(\text{Đặt }y=x+1\Rightarrow y-\frac{1}{2}=x+\frac{1}{2}\text{ ta được:}\)

\(\frac{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}{y^2}=\frac{y^2-y+\frac{1}{4}+\frac{3}{4}}{y^2}\)

\(=\frac{y^2-y+1}{y^2}=\frac{\left(y^2-y+1\right):y^2}{y^2:y^2}=1-\frac{1}{y}+\frac{1}{y^2}\)

\(=\frac{1}{4}-2.\frac{1}{y}.\frac{1}{2}+\frac{1}{y^2}+\frac{3}{4}\)

\(=\left(\frac{1}{y}-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vậy GTNN của \(\frac{x^2+x+1}{\left(x+1\right)^2}\) là 3/4 tại: 

\(\frac{1}{y}=\frac{1}{2}\Rightarrow y=2\Rightarrow x=y-1=2-1=1\)

Answer 2

[Cánh Cụt Vui Vẻ]

\(\frac{3}{4}\) tick nha