ĐKXĐ: \(x\ne0\)
a)\(A=\dfrac{\left(x^2+2x+5\right)x}{2x}=\dfrac{x^2+2x+5}{2}=\dfrac{x^2+2x+1+4}{2}=\dfrac{\left(x+1\right)^2+4}{2}\)
Vì \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow A=\dfrac{\left(x+1\right)^2+4}{2}\ge\dfrac{0+4}{2}=2\)
\(A=2\Leftrightarrow x=-1\)
-Vậy \(A_{min}=2\)
b) ĐKXĐ: \(x\ne2\)
\(B=\dfrac{8-x^3}{3x-6}=\dfrac{-\left(x-2\right)\left(x^2+2x+4\right)}{3\left(x-2\right)}=\dfrac{-\left(x^2+2x+4\right)}{3}=\dfrac{-\left(x^2+2x+1+3\right)}{3}=\dfrac{-\left(x+1\right)^2-3}{3}\)
-Vì \(-\left(x+1\right)^2\le0\forall x\)
\(\Rightarrow B=\dfrac{-\left(x+1\right)^2-3}{3}\le\dfrac{0-3}{3}=-1\)
\(B=-1\Leftrightarrow x=-1\)
-Vậy \(B_{max}=-1\)
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