Question

Answer 1

\(C=-\dfrac{3}{4}x^2+\dfrac{5}{4}x+4=-\dfrac{3}{4}\left(x-\dfrac{5}{6}\right)^2+\dfrac{217}{48}\le\dfrac{217}{48}\)

\(C_{max}=\dfrac{217}{48}\) khi \(x=\dfrac{5}{6}\)

\(D=\dfrac{15}{\left(x-1\right)^2+3}\le\dfrac{15}{3}=5\)

\(D_{max}=5\) khi \(x=1\)

Biểu thức E chỉ có min, không có max

\(E=1-\dfrac{2}{x}+\dfrac{2015}{x^2}=2015\left(\dfrac{1}{x}-\dfrac{1}{2015}\right)^2+\dfrac{2014}{2015}\ge\dfrac{2014}{2015}\)