Question

Tìm giá trị nhỏ nhất của :

\(\frac{x^2+4x+12}{\left(x+1\right)^2}\)

Answer 1

\(\frac{x^2+4x+12}{\left(x+1\right)^2}=\frac{x^2+2x+1}{\left(x+1\right)^2}+\frac{x^2+2x+1}{\left(x+1\right)^2}-\frac{x^2-10}{\left(x+1\right)^2}=2-\frac{x^2-10}{\left(x+1\right)^2}\ge2\)

Dấu " = " xảy ra khi :

\(\frac{x^2-10}{\left(x+1\right)^2}=0\Rightarrow x^2-10=0\Rightarrow x=\sqrt{10}\)

Answer 2

\(A=\frac{x^2+4x+12}{\left(x+1\right)^2}=\frac{9\left(x^2+4x+12\right)}{9\left(x+1\right)^2}=\frac{9x^2+36x+108}{9\left(x+1\right)^2}=\frac{8x^2+16x+8+x^2+20x+100}{9\left(x+1\right)^2}\)

\(A=\frac{8\left(x+1\right)^2+\left(x+10\right)^2}{9\left(x+1\right)^2}=\frac{8}{9}+\frac{\left(x+10\right)^2}{9\left(x+1\right)^2}\ge\frac{8}{9}\)

\(\Rightarrow A_{min}=\frac{8}{9}\) khi \(x+10=0\Rightarrow x=-10\)

Answer 3

vẫn chưa hiểu lắm ''-''