đặt \(A=x\sqrt{6-x}+\left(5-x\right)\sqrt{x+1}\)
\(A=\sqrt{x}\sqrt{x\left(6-x\right)}+\sqrt{5-x}\sqrt{\left(5-x\right)\left(x+1\right)}\)
Áp dụng BĐT bunyakovsky :
\(A^2\le\left(x+5-x\right)\left[x\left(6-x\right)+\left(5-x\right)\left(x+1\right)\right]\)
\(A^2\le5\left(-2x^2+10x+5\right)=5\left[-2\left(x-\frac{5}{2}\right)^2+\frac{35}{2}\right]\)
\(A^2\le\frac{5.35}{2}=\frac{175}{2}=87,5\Leftrightarrow A\le\sqrt{87,5}\)
dấu = xảy ra khi \(\left\{\begin{matrix}x=\frac{5}{2}\\\frac{1}{6-x}=\frac{1}{x+1}\end{matrix}\right.\)<=> x=2,5
vậy Amax=.....
Đúng 0
Bình luận (0)
