\(C=\dfrac{-1}{\left(x-6\sqrt{x}+9\right)+1}=\dfrac{-1}{\left(\sqrt{x}-3\right)^2+1}\ge\dfrac{-1}{1}=-1\)
\(C_{min}=-1\) khi \(x=9\)
\(E=\dfrac{2}{-4-\left(x-2\sqrt{x}+1\right)}=\dfrac{2}{-4-\left(\sqrt{x}-1\right)^2}\ge\dfrac{2}{-4}=-\dfrac{1}{2}\)
\(E_{min}=-\dfrac{1}{2}\) khi \(x=1\)
Đúng 1
Bình luận (0)
