\(A=a^3-b^3-ab\)

   \(=\left(a-b\right)\left(a^2+ab+b^2\right)-ab\)

   \(=a^2+ab+b^2-ab\) (vì \(a-b=1\))

   \(=a^2+b^2\)

   \(=a^2+\left(a-1\right)^2\)

   \(=2a^2-2a+1\)

  \(=2\left(a^2-a+\frac{1}{4}\right)+\frac{1}{2}\)

  \(=2\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall a\)

Dấu "=" xảy ra: \(\Leftrightarrow a-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{2}\)

\(b=a-1=\frac{1}{2}-1=-\frac{1}{2}\)

Vậy \(A_{min}=\frac{1}{2}\Leftrightarrow a=\frac{1}{2},b=-\frac{1}{2}\)

Chúc bạn học tốt.

a) \(x^3+x^2+x=-\frac{1}{3}\)

\(\Leftrightarrow3x^3+3x^2+3x=-1\)

\(\Leftrightarrow2x^3+\left(x^3+3x^2+3x+1\right)=0\)

\(\Leftrightarrow2x^3=-\left(x+1\right)^3\)

\(\Leftrightarrow x.\sqrt[3]{2}=-x-1\)

\(\Leftrightarrow x+x.\sqrt[3]{2}=-1\)

\(\Leftrightarrow x\left(1+\sqrt[3]{2}\right)=-1\)

\(\Leftrightarrow x=\frac{-1}{1+\sqrt[3]{2}}\)

b) \(x^3+2x^2+4x=-\frac{8}{3}\)

\(\Leftrightarrow3x^3+6x^2+12x=-8\)

\(\Leftrightarrow2x^3+\left(x^3+6x^2+12x+8\right)=0\)

\(\Leftrightarrow2x^3=-\left(x+2\right)^3\)

\(\Leftrightarrow x.\sqrt[3]{2}=-x-2\)

\(\Leftrightarrow x\left(1+\sqrt[3]{2}\right)=-2\)

\(\Leftrightarrow x=-\frac{2}{1+\sqrt[3]{2}}\)

\(\sqrt[3]{x-5}+\sqrt[3]{2x-1}-\sqrt[3]{3x+2}=-2\)

\(\Leftrightarrow\sqrt[3]{x-5}+\sqrt[3]{2x-1}+\sqrt[3]{-3x-2}=-2\)

Đặt \(a=\sqrt[3]{x-5}\);\(b=\sqrt[3]{2x-1}\);\(c=\sqrt[3]{-3x-2}\)

\(\Rightarrow\hept{\begin{cases}a+b+c=-2\\a^3+b^3+c^3=-8\end{cases}}\)

\(\Rightarrow\left(a+b+c\right)^3-a^3-b^3-c^3=0\)

\(\Leftrightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

Đến đây bạn tự giải tiếp

\(A+1=\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

\(=\left(ab+bc+ca\right)\left(a+b+c\right)-abc+abc\)

\(=\left(ab+bc+ca\right)\left(a+b+c\right)\)

\(\ge\left(a+b+c\right).3\sqrt[3]{a^2b^2c^2}=3\left(a+b+c\right)\)            Do   abc=1