Question

[Shiro Nerin]

Tim GTNN: A = x^2 + x + 1 / ( x + 1 )^2

Answer 1

[Nguyệt]

\(A=\frac{x^2+2x+1-x-1+1}{\left(x+1\right)^2}=\frac{\left(x+1\right)^2}{\left(x+1\right)^2}-\frac{\left(x+1\right)}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\)

\(A=1-\frac{1}{x+1}+\left(\frac{1}{x+1}\right)^2\)

Đặt B=\(\frac{1}{x+1}\). ta có: 

\(A=B^2-B+1=B^2-\frac{2B.1}{2}+\frac{1}{4}+\frac{3}{4}=\left(B-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

dấu = xảy ra khi \(B-\frac{1}{2}=0\)

\(\Rightarrow B=\frac{1}{2}\). Vậy Min A=\(\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)

Answer 2

[Nguyệt]

eei, sorry :>

\(B=\frac{1}{x+1}=\frac{1}{2}\Rightarrow x+1=2\Rightarrow x=1\)

=.=" sorry bn nha, t làm lộn