Question

tìm GTLN của A=\(\frac{10x}{\left(x+2\right)^2}\)

Answer 1

Ta có:\(\frac{5}{4}-A=\frac{5}{4}-\frac{10x}{\left(x+2\right)^2}=\frac{5\left(x+2\right)^2-40x}{4\left(x+2\right)^2}=\frac{5\left(x^2+4x+4\right)-40x}{4\left(x+2\right)^2}\)

=\(=\frac{5x^2+20x+20-40x}{4\left(x+2\right)^2}=\frac{5x^2-20x+20}{4\left(x+2\right)^2}=\frac{5\left(x^2-4x+4\right)}{4\left(x+2\right)^2}=\frac{5\left(x-2\right)^2}{4\left(x+2\right)^2}\ge0\)

\(\Rightarrow\frac{5}{4}-A\ge0\Rightarrow\frac{5}{4}\ge A\).Nên GTLN của A la \(\frac{5}{4}\) đạt được khi \(x=2\)