Bài IV:
Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\)
\(\Leftrightarrow\left(x^2-5x+2x-10\right)\left(x^2-6x+3x-18\right)=180\)
\(\Leftrightarrow\left(x^2-3x-10\right)\left(x^2-3x-18\right)-180=0\)
\(\Leftrightarrow\left(x^2-3x\right)^2-28\left(x^2-3x\right)+180-180=0\)
\(\Leftrightarrow\left(x^2-3x\right)^2-28\left(x^2-3x\right)=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(x^2-3x-28\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x-7\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x-7=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=7\\x=-4\end{matrix}\right.\)
Vậy: S={0;3;7;-4}
