Bài 1:
a) Ta có: \(a+5⋮a-2\)
\(\Leftrightarrow a-2+7⋮a-2\)
mà \(a-2⋮a-2\)
nên \(7⋮a-2\)
\(\Leftrightarrow a-2\inƯ\left(7\right)\)
\(\Leftrightarrow a-2\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow a\in\left\{3;1;9;-5\right\}\)(tm)
Vậy: \(a\in\left\{3;1;9;-5\right\}\)
b) Ta có: \(3a⋮a-1\)
\(\Leftrightarrow3a-3+3⋮a-1\)
mà \(3a-3=3\left(a-1\right)⋮a+1\)
nên \(3⋮a-1\)
\(\Leftrightarrow a-1\inƯ\left(3\right)\)
\(\Leftrightarrow a-1\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow a\in\left\{2;0;4;-2\right\}\)(tm)
Vậy: \(a\in\left\{2;0;4;-2\right\}\)
c) Ta có: \(5a-8⋮a-4\)
\(\Leftrightarrow5a-20+12⋮a-4\)
mà \(5a-20=5\left(a-4\right)⋮a-4\)
nên \(12⋮a-4\)
\(\Leftrightarrow a-4\inƯ\left(12\right)\)
\(\Leftrightarrow a-4\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
hay \(a\in\left\{5;3;6;2;7;1;8;0;10;-2;16;-8\right\}\)(tm)
Vậy: \(a\in\left\{5;3;6;2;7;1;8;0;10;-2;16;-8\right\}\)
d) Ta có: \(a^2+a+2⋮a+1\)
\(\Leftrightarrow a^2+2a+1-a+1⋮a+1\)
\(\Leftrightarrow\left(a+1\right)^2-a+1⋮a+1\)
mà \(\left(a+1\right)^2⋮a+1\)
nên \(-a+1⋮a+1\)
\(\Leftrightarrow-\left(a-1\right)⋮a+1\)
hay \(a-1⋮a+1\)
\(\Leftrightarrow a+1-2⋮a+1\)
mà \(a+1⋮a+1\)
nên \(-2⋮a+1\)
\(\Leftrightarrow a+1\inƯ\left(-2\right)\)
\(\Leftrightarrow a+1\in\left\{1;-1;2;-2\right\}\)
hay \(a\in\left\{0;-2;1;-3\right\}\)
Vậy: \(a\in\left\{0;-2;1;-3\right\}\)
Bài 2:
a) Ta có: \(4n-3⋮n\)
Vì 4n⋮n
nên -3⋮n
hay n∈Ư(-3)
⇔n∈{1;-1;3;-3}(tm)
Vậy: n∈{1;-1;3;-3}
b) Ta có: -13 là bội của 2n-1
⇔2n-1∈Ư(-13)
⇔2n-1∈{1;-1;13;-13}
⇔2n∈{2;0;14;-12}
hay n∈{1;0;7;-6}(tm)
Vậy: n∈{1;0;7;-6}
Bài 1:
a/ a + 5 ⋮ a - 2
=> a - 2 + 7 ⋮ a - 2
Có a - 2 ⋮ a - 2 nên để a + 5 ⋮ a - 2 thì 7 ⋮ a - 2
=> a - 2 ∈ Ư(7)
=> a - 2 ∈ {1; -1; 7; -7}
=> a ∈ {3; 1; 9; -5}
b/ 3a ⋮ a - 1
=> 3. (a + 1) + 3 ⋮ a - 1
Có: 3. (a + 1) ⋮ a - 1 => Để 3a ⋮ a - 1 thì 3 ⋮ a - 1
=> a - 1 ∈ Ư(3)
=> a - 1 ∈ {1; -1; 3; -3}
=> a ∈ {2; 0; 4; -2}
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