17A
a, Ta có: 2x+3$\vdots$x
⇒x∈Ư(3)={±1;±3}
Vậy x∈{±1;±3}
b, Ta có: 8x+4$\vdots$2x-1
⇒4(2x-1)+8$\vdots$2x-1
⇒2x-1∈Ư(8)={±1;±2;±4;±8}
2x-1=1⇒x=1 (tm)
2x-1=-1⇒x=0 (tm)
2x-1=2⇒x=3/2 (laoij)
2x-1=-2⇒x=-1/2 (loại)
2x-1=4⇒x=5/2 (loại)
2x-1=-4⇒x=-3/2 (laoij)
2x-1=8⇒x=9/2 (loại)
2x-1=-8⇒x=-7/2 (loại)
Vậy x∈{0;1}
c, Ta có: x²-5x+7$\vdots$x-5
⇒x(x-5)+7$\vdots$x-5
⇒x-5∈Ư(7)={±1;±7}
x-5=1⇒x=6
x-5=-1⇒x=4
x-5=7⇒x=12
x-5=-7⇒x=-2
Vậy x∈{6;4;12;-2}
7B
a, Ta có: x+3$\vdots$x
⇒x∈Ư(3)={±1;±3}
Vậy x∈{±1;±3}
b, Ta có: 6x+4$\vdots$2x-1
⇒3(2x-1)+7$\vdots$2x-1
⇒2x-1∈Ư(7)={±1;±7}
2x-1=1⇒x=1
2x-1=-1⇒x=0
2x-1=7⇒x=4
2x-1=-7⇒x=-3
Vậy x∈{1;0;4;-3}
c, Ta có: n²-x+7$\vdots$x-1
⇒x(x-1)+7$\vdots$x-1
⇒x-1∈Ư(7)={±1;±7}
x-1=1⇒x=2
x-1=-1⇒x=0
x-1=7⇒x=8
x-1=-7⇒x=-6
Vậy x∈{2;0;8;-6}
