Question

Tìm MAX A = x² / x²+2x + 2010

Answer 1

\(A=\frac{x^2}{x^2+2x+2010}\)

- Với \(x=0\Rightarrow A=0\)

- Với \(x\ne0\Rightarrow A=\frac{1}{1+\frac{2}{x}+\frac{2010}{x^2}}=\frac{1}{2010\left(\frac{1}{x^2}+2.\frac{1}{2010}.\frac{1}{x}+\frac{1}{2010^2}\right)+1-\frac{1}{2010}}\)

\(A=\frac{1}{2010\left(\frac{1}{x}+\frac{1}{2010}\right)^2+\frac{2009}{2010}}\le\frac{1}{\frac{2009}{2010}}=\frac{2010}{2009}\)

So sánh \(0\) và \(\frac{2010}{2009}\)

\(\Rightarrow A_{max}=\frac{2010}{2009}\) khi \(\frac{1}{x}+\frac{1}{2010}=0\Rightarrow x=-2010\)