tìm x ϵ Z sao cho
2x+3 chia hết cho x-1
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a) 6 ⋮ x
⇒ x ∈ Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}
b) 8 ⋮ (x + 1)
⇒ x + 1 ∈ Ư(8) ={-8; -4; -2; -1; 1; 2; 4; 8}
⇒ x ∈ {-9; -5; -3; -2; 0; 1; 3; 7}
3/ Ta có:
\(A=\dfrac{1-2x}{x+3}\)
\(A=\dfrac{-2x+1}{x+3}\)
\(A=\dfrac{-2x-6+7}{x+3}\)
\(A=\dfrac{-2\left(x+3\right)+7}{x+3}\)
\(A=-2+\dfrac{7}{x+3}\)
A nguyên khi \(\dfrac{7}{x+3}\) nguyên
⇒ 7 ⋮ \(x+3\)
\(\Rightarrow x+3\inƯ\left(7\right)\)
\(\Rightarrow x+3\in\left\{1;-1;7;-7\right\}\)
\(\Rightarrow x\in\left\{-2;-4;4;-10\right\}\)
a, Vì : \(6⋮x-1\Rightarrow x-1\inƯ\left(6\right)\)
Mà : \(Ư\left(6\right)=\left\{1;2;3;6\right\}\Rightarrow x\in\left\{2;3;4;7\right\}\)
Vậy ...
b,Vì : \(14⋮2x+3\Rightarrow2x+3\inƯ\left(14\right)\)
Mà : \(Ư\left(14\right)=\left\{1;2;7;14\right\}\) ; \(2x+3\ge3\Rightarrow2x+3\in\left\{7;14\right\}\)
Ta có : 2x + 3 là số lẻ
=> 2x + 3 = 7
=> 2x = 4 => x = 2
Vậy x = 2
c, \(x-1⋮12\Rightarrow x-1\in B\left(12\right)\)
Mà : \(B\left(12\right)=\left\{0;12;24;36;...\right\}\) ; 0 < x < 30
\(\Rightarrow x-1\in\left\{12;24\right\}\)
\(\Rightarrow x\in\left\{13;25\right\}\)
Vậy ...
x + 7 ⋮ x + 1
ta có : x + 7 = x + 1 + 6
nên x + 1 + 6 ⋮ x + 1 và x + 1 ⋮ x + 1
<=> 6 ⋮ x + 1
<=> x + 1 = { 1; 2; 3; 6 }
XÉT CÁC TRƯỜNG HỢP
\(\cdot\) nếu x + 1 = 1 thì suy ra x = 0 (TM)
\(\cdot\) nếu x + 1 = 2 thì suy ra x = 1 (TM)
\(\cdot\) nếu x + 1 = 3 thì suy ra x = 2 (TM)
\(\cdot\) nếu x + 1 = 6 thì suy ra x = 5 (TM)
vậy x = {0; 1; 2; 5}
a) \(25⋮n+2\left(n\in Z\right)\)
\(\Rightarrow n+2\in\left\{-1;1;-5;5;-25;25\right\}\)
\(\Rightarrow n\in\left\{-3;-1;-7;3;-27;23\right\}\)
b) \(2n+4⋮n-1\)
\(\Rightarrow2n+4-2\left(n-1\right)⋮n-1\)
\(\Rightarrow2n+4-2n+2⋮n-1\)
\(\Rightarrow6⋮n-1\)
\(\Rightarrow n-1\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
\(\Rightarrow n\in\left\{0;2;-1;3;-2;4;-5;7\right\}\)
c) \(1-4n⋮n+3\)
\(\Rightarrow1-4n+4\left(n+3\right)⋮n+3\)
\(\Rightarrow1-4n+4n+12⋮n+3\)
\(\Rightarrow13⋮n+3\)
\(\Rightarrow n+3\in\left\{-1;1;-13;13\right\}\)
\(\Rightarrow n\in\left\{-4;-2;-15;10\right\}\)
a) n ϵ{−3;−1;−7;3;−27;23}
b) n ∈{0;2;−1;3;−2;4;−5;7}
c) n ϵ {−4;−2;−15;10}
2x + 3 ⋮ x - 1
⇒ 2x - 2 + 5 ⋮ x - 1
2(x-1) + 5 ⋮ x - 1
5 ⋮ x - 1
x - 1 \(\in\) { -5; -1; 1; 5}
x \(\in\) { -4; 0; 2; 6}