Ta có :\(\frac{2}{-4x^2+8x-5}=\frac{-2}{4x^2-8x+5}\)
\(=\frac{-2}{\left(4x^2-8x+16\right)-11}\)
\(=\frac{-2}{\left(2x-16\right)^2-11}\ge\frac{-2}{-11}\)
\(=\frac{-2}{\left(2x-4\right)^2-11}\ge\frac{2}{11}\)
Dấu = xảy ra khi :(2x-4)2 = 0
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow x=2\)
Vậy gtnn của bt là \(\frac{2}{11}\Leftrightarrow x=2\)
C= \(\frac{2}{-4x^2+8x-5}\)
\(\Rightarrow\frac{-2}{4x^2-8x+5}\)
\(\Rightarrow\frac{-2}{\left(4x^2-8x+4\right)+1}\)
\(\Rightarrow\frac{-2}{\left(2x-x\right)^2+1}\)
⇒(2x-2)2 ≥ 0
⇒ (2x-2)2+1 ≥ 1 > 0
\(\Rightarrow\frac{-2}{\left(2x-2\right)^2+1}\ge\frac{-2}{1}\)
\(\Rightarrow C\ge\frac{-2}{1}\)
\(\Rightarrow\frac{1}{\left(2x-2\right)^2+1}\le1\)
Dấu = xảy ra ↔ 2x-2 =0 ⇒ x=1
Vậy GTNN của C=1 khi x=1