a ) Vì `30 \vdots x`

`=> x in Ư(30)= {-30 ; -15 ; -10 ; -6 ; -5 ; -3 ;- 2 ; -1 ; 1 ;2 ; 3 ; 5 ;6 ; 10 ; 15 ; 30}`

Vì `x < 8` nên `x in {-30 ; -15 ; -10 ; -6 ; -5 ; -3 ; -1 ; -1 ; 2; 3;5;6}`

Vậy `x in {-30 ; -15 ; -10 ; -6 ; -5 ; -3 ; -1 ; -1 ; 2; 3;5;6}`

c ) Vì `90 \vdots x   ;   126 \vdots x` nên `x in ƯC(90 ;126)`

Ta có :

`90 = 2 . 3^2 . 5`

`126 = 2 . 3^2 . 7`

`=> ƯCLN(90 ; 126) = 2. 3^2 = 18`

`=> x in Ư(18) = {-18;-9;-6;-3;-2;-1;1;2;3;6;9;18}`

Vì `x > 9` nên `x = 18`

Vậy `x = 18`

e ) Vì `x \vdots 35   ;    x \vdots 63   ;   x \vdots 105` nên `x in BC(35 ; 63 ; 105)`

Ta có :

`35 = 5.7`

`63 = 3^2 . 7`

`105 = 3.5.7`

`=> BCNNN(35 ; 63 ; 105) = 3^2 . 5 . 7 = 315`

`=> x in B(315) = {0 ; 315 ; 630 ; 945 ; ...}`

Vì `315 < x < 632` nên `x = 630`

Vậy `x= 630`

`#dtkc`

`a)`

`30 \vdots x`

`=> x in Ư(30) = {1; -1; 2; -2; 3; -3; 5; -5; 6; -6; 10; -10; 15; -15; 30; -30}`

Vì `x < 8`

`=> x in {1; -1; 2; -2; 3; -3; 5; -5; 6; -6; -10; -15; -30}`

`c)`

`90 \vdots x`

`126 \vdots`

`=> x in ƯC(90; 126) `

`ƯCLN(90; 126) = 2 . 3^2 = 18`

`=> ƯC(90; 126) = {1; -1; 2; -2; 3; -3; 6; -6; 9; -9; 18; -18}`

Vì `x > 9`

`=> x = 18`

`e)`

`x \vdots 35`

`x \vdots 63`

`x \vdots 105`

`=> x in BC(35; 63; 105) `

`BCNNNN{35; 63; 105) = 3^2 . 5 . 7 = 315`

`=> BC(35; 63; 105) = {0; ±315; ±630}`

Vì `315 < x < 632`

`=> x = 630`