Question

Tim so nguyen a de 2a+1chia het a-1

Answer 1

\(2a+1⋮a-1\)

\(2a-2+3⋮a-1\)

\(2\left(a-1\right)+3⋮a-1\)

\(\Rightarrow\left\{{}\begin{matrix}2\left(a-1\right)⋮a-1\\3⋮a-1\end{matrix}\right.\)

\(3⋮a-1\)

\(\Rightarrow a-1\in\)Ư\(\left(3\right)=\left\{-3;-1;1;3\right\}\)

a-1	-3	-1	1	3
a	-2	0	2	4

\(\Rightarrow a\in\left\{-2;0;2;4\right\}\)

 

Answer 2

Ta có: \(2a+1⋮a-1\)

\(\Leftrightarrow2a-2+3⋮a-1\)

mà \(2a-2⋮a-1\)

nên \(3⋮a-1\)

\(\Leftrightarrow a-1\inƯ\left(3\right)\)

\(\Leftrightarrow a-1\in\left\{1;-1;3;-3\right\}\)
hay \(a\in\left\{2;0;4;-2\right\}\)

Vậy: \(a\in\left\{2;0;4;-2\right\}\)

Answer 3

2a+1⋮a−1

2a−2 2(a−1)+3⋮a−1

⇒⎨2(a−1)⋮a−1

      3⋮a−1

Answer 4

Giải:

2a+1 ⋮ a-1

⇒2a-2+3 ⋮ a-1 

⇒3 ⋮ a-1

⇒a-1 ∈ Ư(3)={-3;-1;1;3}

Ta có bảng:

a-1=-3 ➜a=-2

a-1=-1 ➜a=0

a-1=1 ➜a=2

a-1=3 ➜a=4

Vậy a ∈ {-2;0;2;4}

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