Question

Giúp mình Late 2 với mình đang cần gấp

Answer 1

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