b) Ta có: (x-7).(x+3) <0
\(\Rightarrow\) có 1 số là số nguyên dương, 1 số là số nguyên âm.
Mà x+3>x-7 \(\Rightarrow\left\{{}\begin{matrix}x+3>0\\x-7< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x< 7\end{matrix}\right.\)
\(\Rightarrow-3< x< 7\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
Vậy:........
/ là trị tuyệt đối à?
a,
\(\left|5x-2\right|\le0\\ Vì\left|A\right|\ge0\\ \Rightarrow\left|5x-2\right|=0\\ \Leftrightarrow5x-2=0\\ \Leftrightarrow5x=2\\ \Leftrightarrow x=\dfrac{2}{5}\)
b,
\(\left(x-7\right)\left(x+3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\left(VL\right)}\\\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\end{matrix}\right.\\ \Rightarrow-3< x< 7\)
\(a)\left|5x-2\right|\le0\)
Vì vế trái luôn luôn \(\ge0\) nên khẳng định này chỉ đúng khi:
\(\left|5x-2\right|=0\\ \Leftrightarrow x=\dfrac{2}{5}\)
\(b)\left(x-7\right)\left(x+3\right)< 0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
