giải nhanh giúp em vs ak

Đáp án:

Giải thích các bước giải:

 `25`

`a`

`-13 vdots x+1`

`to x+1 in Ư_(13)={-13;-1;1;13}`

`to x in {-14;-2;0;12}`

Vậy `...`

`b`

`x+5 vdots x-2`

`to x-2+7 vdots x-2`

`to 7 vdots x-2`

`to x-2 in Ư_((7))={-7;-1;1;7}`

`to x on {-5;1;3;9}`

Vậy `....`

`c`

`3x-8 vdots x-4`

`to 3(x-4) +4 vdots x-4`

`to 4 vdots x-4`

`to x-4 in Ư_((4))={-4;-1;1;4}`

`to x in {0;3;5;8}`

Vậy `...`

`d`

`x^2-x-1 vdots x-1`

`to x(x-1)-1 vdots x-1`

`to 1 vdots x-1`

`to x-1 in Ư_((1))={-1;1}`

`to x in {0;2}`

Vậy `...`

`26`

`a`

`(x-3)(y+4)=6`

`(x-3)(y-4)=1*6=2*3=3*2=6*1=(-1)*(-6)=(-2)*(-3)=(-3)*(-2)=(-6)*(-1)`

`to {x-3;y-4} in {(1;6);(2;3);(3;2);(6;1);(-1;-6);(-2;-3);(-3;-2);(-6;-1)}`

`to (x;y) in {(4;10);(6;7);(6;6);(9;5);(2;-2);(1;1);(0;2);(-3;3)}`

Vậy `...`

`b`

`(x-2)(5y+1)=12`

`to 12 vdots 5y+1`

`to 5y+1 in Ư_((12))`

Mà `5y+1` là số lẻ nên

`5y+1 in {-3;-1;1;3}`

`TH1:`

`5y+1=-3 to 5y=-4 to y=-4/5 (L)`

`TH2:`

`5y+1=-1 to 5y=-2 to y=-2/5 (L)`

`TH3:`

`5y+1 =1 to 5y=0 to y=0 (tm)`

`to x-2 = 12 to x=14`

`TH4:`

`5y+1=3 to 5y=2 to y=5/2 (L)`

Vậy `x=14;y=0`