Question

Tìm x, biết:

Q= \(\dfrac{x^2+x+1}{x^2+2x+1}\) đạt Min

Answer 1

Lời giải:
\(A=\frac{x^2+x+1}{x^2+2x+1}=\frac{x^2+2x+1-x}{x^2+2x+1}=1-\frac{x}{x^2+2x+1}=1-\frac{x}{(x+1)^2}\)

Ta thấy \((x+1)^2-4x=x^2-2x+1=(x-1)^2\geq 0\)

\(\Rightarrow (x+1)^2\geq 4x\Rightarrow \frac{x}{(x+1)^2}\leq \frac{x}{4x}=\frac{1}{4}\)

\(\Rightarrow A=1-\frac{x}{(x+1)^2}\geq 1-\frac{1}{4}=\frac{3}{4}\)

Vậy \(A_{\min}=\frac{3}{4}\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1\), tức là A đạt min khi $x=1$