a/ \(\left|x+1\right|>2\)
\(\Rightarrow x+1\in\left\{3;-3;...\right\}\)
\(\Rightarrow x\in\left\{2;-4;...\right\}\)
Vậy...
b) \(\left|\dfrac{1}{3}x\right|.\left|-2,7\right|=\left|-9\right|\)
\(\Rightarrow\left|\dfrac{1}{3}x\right|.2,7=9\)
\(\Rightarrow\left|\dfrac{1}{3}x\right|=9:2,7=\dfrac{10}{3}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{10}{3}\\\dfrac{1}{3}x=\dfrac{-10}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\).
Vậy \(x=10;x=-10.\)
\(\left|x+1\right|>2\)
Ta có: gttđ của 1 số luôn \(\ge\)0
mà \(\left|x+1\right|>2\)
\(x+1=\left\{-5;-4;-3;.......;3;4;5;..........\right\}\)
\(x=-6;-5;-4;.................;2;3;4;.....\)
b)\(\left|\dfrac{1}{3}x\right|.\left|-2,7\right|=\left|-9\right|\)
\(\left|\dfrac{1}{3}x\right|.2,7=9\)
\(\left|\dfrac{1}{3}x\right|=9:2,7=\dfrac{10}{3}\)
\(\dfrac{1}{3}x=\dfrac{10}{3}\Rightarrow x=10\)
\(\dfrac{1}{3}x=\dfrac{-10}{3}\Rightarrow x=-10\)
T nghĩ đề nên cho thêm ĐK x ϵ Z thì ý a ms lm giống như bn Ngọc Anh đc, nếu k có Đk thì:
\(a\)) \(\left|x+1\right|>2\)
\(\Rightarrow\left[{}\begin{matrix}x+1>2\\x+1< -2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x>1\\x< -3\end{matrix}\right.\)
Vậy để \(\left|x+1\right|>2\) thì \(x>1\) hoặc \(x< -3\)