a: \(A=\dfrac{\left(x^2+2x+5\right)\cdot x}{2\cdot x}=\dfrac{x^2+2x+5}{2}=\dfrac{\left(x+1\right)^2+4}{2}\ge2\forall x\)
Dấu '=' xảy ra khi x=-1
b: \(B=\dfrac{8-x^3}{3x-6}=\dfrac{-\left(x-2\right)\left(x^2+2x+4\right)}{3\left(x-2\right)}\)
\(=\dfrac{-\left(x^2+2x+4\right)}{3}=\dfrac{-\left(x+1\right)^2-3}{3}\le-1\)
Dấu '=' xảy ra khi x=-1
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