\(a.-4\left(x-5\right)< 0\)
\(\Leftrightarrow x-5>0\)
\(\Leftrightarrow x>5\)
\(b.\left(x-2\right)\left(x-6\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x-6< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x-6>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2< x< 6\left(n\right)\\6< x< 2\left(l\right)\end{matrix}\right.\)
a) Ta có: \(-4\cdot\left(x-5\right)< 0\)
\(\Leftrightarrow-4\) và x-5 khác dấu
mà -4<0
nên x-5>0
hay x>5
Vậy: x>5
b) Ta có: \(\left(x-2\right)\left(x-6\right)< 0\)
\(\Leftrightarrow x-2;x-6\) khác dấu
Trường hợp 1:
\(\left\{{}\begin{matrix}x-2>0\\x-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< 6\end{matrix}\right.\Leftrightarrow2< x< 6\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x-2< 0\\x-6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>6\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy: 2<x<6
b, (x - 2)(x - 6) < 0
\(\Rightarrow\) Xét 2 TH:
TH1: \(\left[{}\begin{matrix}x-2< 0\\x-6>0\end{matrix}\right.\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x< 2\\x>6\end{matrix}\right.\)
\(\Rightarrow\) 2 > x > 6 (Không có giá trị nào thỏa mãn x \(\in\) Z)
TH2: \(\left[{}\begin{matrix}x-2>0\\x-6< 0\end{matrix}\right.\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x>2\\x< 6\end{matrix}\right.\)
\(\Rightarrow\) 2 < x < 6
Vì x \(\in\) Z nên x \(\in\) {3; 4; 5}
Vậy x \(\in\) {3; 4; 5}
a, -4(x - 5) < 0
\(\Rightarrow\) x - 5 > 0 (Vì -4 < 0)
\(\Rightarrow\) x > 5
Vậy x \(\in\) Z thỏa mãn x > 5
Chúc bn học tốt!