Question

giải bất phương trình

(x^3-27)(x^3-1)(2x+3-x^2) >=0

Answer 1

Ta có: \(\left(x^3-27\right)\left(x^3-1\right)\left(2x+3-x^2\right)\ge0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)\left(x-1\right)\left(x^2+x+1\right)\left[-\left(x^2-2x-3\right)\right]\ge0\)

Vì \(\left\{{}\begin{matrix}x^2+3x+9=\left(x+\frac{3}{2}\right)^2+\frac{27}{4}>0\left(đúng\right)\\x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(đúng\right)\\-\left(x^2-2x-3\right)=-\left(x-1\right)^2+2\le2\end{matrix}\right.\)

Nên : \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x>3\end{matrix}\right.\left(vl\right)}\end{matrix}\right.\Rightarrow1< x< 3}\)