Question

Tìm \(k\inℝ\)sao cho \(\frac{3k}{\left(k+1\right)^2}\)đạt giá trị lớn nhất

Answer 1

Gọi \(A=\frac{3k}{\left(k+1\right)^2}\)

Đặt \(\frac{1}{k+1}=t\Rightarrow k+1=\frac{1}{t}\Rightarrow k=\frac{1}{t}-1\)

Khi đó \(A=\frac{3k}{\left(k+1\right)^2}=3k\cdot\frac{1}{\left(k+1\right)^2}=3\left(\frac{1}{t}-1\right)t^2\)

\(=-3t^2+3t=-3\left(t^2-t\right)=-3\left(t^2-t+\frac{1}{4}\right)+\frac{3}{4}=-3\left(t-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(t-\frac{1}{2}\right)^2\ge0\Rightarrow-3\left(t-\frac{1}{2}\right)^2\le0\Rightarrow A=-3\left(t-\frac{1}{2}\right)^2+\frac{3}{4}\le\frac{3}{4}\)

Dấu "=" xảy ra khi \(t=\frac{1}{2}\Leftrightarrow k=1\)

Vậy Amax = 3/4 khi k=1