Question

Answer 1

\(M=\left(x-3\right)^3+\left(-x-1\right)^3\)

\(=\left(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\right)+\left[\left(-x\right)^3+3\cdot\left(-x\right)^2\cdot\left(-1\right)+3\cdot\left(-x\right)\cdot\left(-1\right)^2+\left(-1\right)^3\right]\)

\(=x^3-9x^2+27x-27-x^3-3x^2-3x-1\)

\(=-12x^2+24x-28\)

\(=-12x^2+24x-12-16\)

\(=-12\left(x^2-2x+1\right)-16\)

\(=-12\left(x-1\right)^2-16\)

Mà: \(-12\left(x-1\right)^2\le0\forall x\)

\(\Rightarrow M=-12\left(x-1\right)^2-16\le-16\forall x\)

Dấu "=" xảy ra:

\(-12\left(x-1\right)^2-16=-16\Rightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy: \(M_{max}=-16\Leftrightarrow x=1\)