Question

[Phạm Thị Thanh Thanh]

tìm m thuộc Z để :

a, giá trị của biểu thức m-1 chia hết cho giá trị của biểu thức 2m+1

b, | 3m-1| < 3

Answer 1

[Nguyễn Thanh Hằng]

a/ Ta có :

\(m-1⋮2m+1\)

Mà \(2m+1⋮2m+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m-2⋮2m+1\\2m+1⋮2m+1\end{matrix}\right.\)

\(\Leftrightarrow3⋮2m+1\)

Vì \(m\in Z\Leftrightarrow2m+1\in Z;2m+1\inƯ\left(3\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2m+1=1\\2m+1=-1\\2m+1=3\\2m+1=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-1\\m=1\\m=-2\end{matrix}\right.\)

Vậy ....

b/ Ta có :

\(\left|3m-1\right|< 3\)

Mà \(\left|3m-1\right|\ge0\)

\(\Leftrightarrow\left|3m-1\right|\in\left\{0;1;2\right\}\)

+) \(\left|3m-1\right|=0\)

\(\Leftrightarrow3m-1=0\)

\(\Leftrightarrow3m=1\)

\(\Leftrightarrow m=\dfrac{1}{3}\)\(\left(loại\right)\)

+) \(\left|3m-1\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}3m-1=1\\3m-1=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3m=2\\3m=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{2}{3}\left(loại\right)\\m=0\left(tm\right)\end{matrix}\right.\)

+) \(\left|3m-1\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}3m-1=2\\3m-1=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3m=3\\3m=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m=1\left(tm\right)\\m=-\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)

Vậy ..

Answer 2

[Linh Nguyễn]

\(m-1⋮2m+1\)

\(\Rightarrow2\left(m-1\right)⋮2m+1\)

\(\Rightarrow2m-2⋮2m+1\)

\(\Rightarrow2m-3+1⋮2m+1\)

\(2m+1⋮2m+1\Rightarrow3⋮2m+1\)

\(\Rightarrow2m+1\inƯ\left(3\right)\)

\(Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2m+1=1\\2m+1=-1\\2m+1=3\\2m+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-1\\m=1\\m=-2\end{matrix}\right.\)

\(\left|3m-1\right|< 3\)

\(\Rightarrow\left[{}\begin{matrix}3m-1< 3\\3m-1>-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m< \dfrac{4}{3}\\m>-\dfrac{2}{3}\end{matrix}\right.\)