\(E=\frac{4}{\left(2x-3\right)^2+5}\)
\(E\le\frac{4}{5}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)
\(E=\frac{4}{\left(2x-3\right)^2+5}\)
\(E\le\frac{4}{5}\forall x\)
Dấu " = " xảy ra <=> 2x - 3 = 0 <=> x = 3/2
\(E=\frac{4}{\left(2x-3\right)^2+5}\le\frac{4}{0+5}=\frac{4}{5}\)
Dấu "=" \(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy ............
\(E=\frac{4}{\left(2x-3\right)^2+5}\le\frac{4}{0+5}=\frac{4}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{3}{2}\)
Vậy ..........