c, Do (x - 3)(7 - x) > 0
=> x - 3; 7 - x cùng dấu
Xét 2 TH :
TH1 : \(\left\{{}\begin{matrix}x-3>0\\7-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\7>x\end{matrix}\right.\)<=> 3 < x < 7 (thỏa mãn)
TH2 : \(\left\{{}\begin{matrix}x-3< 0\\7-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\7< x\end{matrix}\right.\)<=> 7 < x < 3 (vô lý)
Vậy x = 4; 5; 6
d, (x - 3)(x - 7) < 0
=> x - 3; x - 7 khác dấu
Xét 2 TH :
TH1 : \(\left\{{}\begin{matrix}x-3>0\\x-7< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\x< 7\end{matrix}\right.\)<=> 3 < x < 7 (thỏa mãn)
TH2 : \(\left\{{}\begin{matrix}x-3< 0\\x-7>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x>7\end{matrix}\right.\)<=> 7 < x < 3 (vô lý)
Vậy x = 4; 5; 6
@Jenny Jenny
