a, x+3 chia hết cho x-1
Ta có: x+3=(x+1)+2
=> 2 chia hết cho x+1
=>x+1 thuộc Ư(2)= {1, -1, 2, -2}
=> x thuộc {0,-2, 1, -3}
b.
b,3x chia hết cho x-1
c,2-x chia hết cho x+1
Ta có:
\(\dfrac{x+3}{x-1}=\dfrac{x-1+4}{x-1}=1+\dfrac{4}{x-1}\)
Để (x + 3) \(⋮\left(x-1\right)\) thì 4 \(⋮\left(x-1\right)\)
\(\Rightarrow\) x - 1 = 1; x - 1 = -1; x - 1 = 2; x - 1 = -2; x - 1 = 4; x - 1 = -4
*) x - 1 = 1
x = 2
*) x - 1 = -1
x = 0
*) x - 1 = 2
x = 3
*) x - 1 = -2
x = -1
*) x - 1 = 4
x = 5
*) x - 1 = -4
x = -3
Vậy x = 5; x = 3; x = 2; x = 0; x = -1; x = -3
a) Ta có: x + 3 \(⋮\)t x - 1
\(\Rightarrow\) (x - 1) + 4 \(⋮\) x - 1
do x - 1 \(⋮\) x-1
\(\Rightarrow\) 4 \(⋮\) x -1
\(\Rightarrow\) x - 1 \(\in\) Ư(4) = {4;-4;2;-2;1;1}
✳ x - 1 = 4 ✳ x - 1 = -4 ✳ x - 1 = 2
x = 4 + 1 =5 x = -4 + 1 = -3 x = 2 + 1 = 3
✳ x - 1 = -2 ✳ x - 1 = 1 ✳ x - 1 = -1
x = -2 + 1 = 1 x = 1 + 1 = 2 x = -1 + 1 = 0
\(\Rightarrow\) x = {5;-3;3;1;2;0}
c) Ta có: 2 - x \(⋮\) x + 1
⇒ 1 - (x + 1) ⋮ x + 1
do x + 1 ⋮ x + 1
⇒1 ⋮ x +1
⇒ x - 1 ∈ Ư(4) = {-1;1}
✳ x + 1 = 1 ✳ x + 1 = - 1
x = 1 + 1 = 2 x = -1 + 1 = 0
⇒ x = {2;0}