Question

Giải bất phương trình :

\(\sqrt[3]{\left(x-1\right)\left(x+2\right)}+\left(x+4\right)\left(x-3\right)\ge0\)

Answer 1

\(\sqrt[3]{\left(x-1\right)\left(x+2\right)}+\left(x+4\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\sqrt[3]{x^2+x-2}+x^2+x-12\ge0\)

Đặt \(\sqrt[3]{x^2+x-2}=a\)

\(\Leftrightarrow a^3+a-10\ge0\)

\(\Leftrightarrow\left(a-2\right)\left(a^2+2a+5\right)\ge0\)

Dễ thấy \(a^2+2a+5>0\Rightarrow a\ge2\)

\(\Leftrightarrow\sqrt[3]{x^2+x-2}\ge2\)

\(\Leftrightarrow x^2+x-10\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\frac{-1+\sqrt{41}}{2}\\x\le\frac{-1-\sqrt{41}}{2}\end{matrix}\right.\)