Lời giải:
a) \((x-7)(x+3)<0\Leftrightarrow \) \(\left[ \begin{matrix} \left\{\begin{matrix} x-7>0\\ x+3<0\end{matrix}\right.(1)\\ \left\{\begin{matrix} x-7<0\\ x+3>0\end{matrix}\right.(2)\end {matrix}\right.\)
\((1)\Leftrightarrow \left\{\begin{matrix} x>7\\ x< -3\end{matrix}\right.\Leftrightarrow 7< x< -3\) (vô lý)
\((2)\Leftrightarrow \left\{\begin{matrix} x< 7\\ x>- 3\end{matrix}\right.\Leftrightarrow 7> x> -3\)
Vì \(x\in\mathbb{Z}\Rightarrow x\in\left\{-2; -1; 0; 1; 2;3 ;4;5;6\right\}\)
b)
\(S=3^0-3^1+3^2-3^3+...+3^{98}-3^{99}\)
\(S=(3^0-3^1+3^2-3^3)+(3^4-3^5+3^6-3^7)+...+(3^{96}-3^{97}+3^{98}-3^{99})\)
\(S=(3^0-3^1+3^2-3^3)+3^4(3^0-3^1+3^2-3^3)+....+3^{96}(3^0-3^1+3^2-3^3)\)
\(S=(1+3^4+3^8+...+3^{96})(3^0-3^1+3^2-3^3)\)
\(S=-20(1+3^4+3^8+...+3^{96})\)
Do đó S là bội của \(-20\)
Ta có đpcm.
