Ta có: \(-3x^2+9x-40\)
\(=-3\left(x^2-3x+\frac{40}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{133}{12}\right)\)
\(=-3\left(x-\frac{3}{2}\right)^2-\frac{133}{4}\)
Ta có: \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow-3\left(x-\frac{3}{2}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x-\frac{3}{2}\right)^2-\frac{133}{4}\le\frac{-133}{4}\forall x\)
\(\Rightarrow\frac{8}{-3\left(x-\frac{3}{2}\right)^2-\frac{133}{4}}\ge\frac{8}{-\frac{133}{4}}=8:\frac{-133}{4}=8\cdot\frac{4}{-133}=\frac{-32}{133}\)
Dấu '=' xảy ra khi \(x-\frac{3}{2}=0\)
hay \(x=\frac{3}{2}\)
Vậy: Giá trị nhỏ nhất của biểu thức \(L=\frac{8}{-3x^2+9x-40}\) là \(-\frac{133}{4}\) khi \(x=\frac{3}{2}\)
