Bạn ăn luôn yêu cầu rồi à !

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

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

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

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

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

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

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

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

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

Tham khảo câu hỏi tương tự nhé bạn .

Tick tớ đc ko 

4x⁴ + 4x³ + 5x² + 2x +1

= (4x⁴ + 4x³ + x² ) + ( 2x² + 1 ) + 1

= x²(2x + 1 )² + 2x(2x + 1) +1

= (x(2x + 1 ) + 1)²

= (2x + x + 1)²

=x^2-4x+3x-12

=x[x-4]+3[x-4]

=[x+3][x-4]

C2:=x^2-16-[x-4]

     =[x-4][x+4]-[x-4]

    =[x-4][x+4-1]

   =[x-4][x+3]

Cách nữa nè !

x^2-x-12

=(x^2-9)-(x+3)

=(x-3)(x+3)-(x+3)

=(x+3)(x-4)