Question

Tìm x,y thuộc Z thỏa (y+1)(x-2)=7

Answer 1

(y+1)(x-2) = 7

=> y+1 và x-2 \(\in\) Ư(7) = {-1,-7,1,7}

Ta có bảng :

y+1	-1	-7	1	7
x-2	-7	-1	7	1
x	-2	-8	0	6
y	-5	1	9	3

Vậy ta có các cặp x,y thõa mãn : (x=-2,y=-5);(x=-8,y=1);(x=0,y=9);(x=6,y=3)

Answer 2

(y+1)(x-2)=7=1.7=7.1=-1.(-7)=(-7).(-1)

Nếu y + 1 = 1 ; x - 2 = 7 => y = 0; x = 9.

Nếu y + 1 = 7 ; x - 2 = 1 => y = 6 ; x = 3

Nếu y + 1 = -1; x - 2 = -7 => y = -2; x = -5

Nếu y + 1 = -7 ; x - 2 = -1 => y = -8; x = 1.

Answer 3

(y+1).(x-2)=1.7=-1.(-7)

TH1:(y+1).(x-2)=1.7

=>y+1=1=>y=0 €Z(T/m)

x-2=7 =>x=9€Z T/m

​TH2:y+1=7 =>y=6€Z T/m

​x-2=1 => x=3€Z (T/m)

TH3:y+1=-1 =>y=-2€Z T/m TH4:y+1=-7 =>y=-8€Z Tm

​x-2=-7 =>x= -5€Z T/m x-2=-1 =>x=1

Answer 4

Ta có : ( y + 1 ) . ( x - 2 ) = 7 = 7 . 1 = 1 . 7 = ( - 1 ) . ( - 7  ) = ( - 7 ) . ( - 1 )

TH1: ( y + 1 ) . ( x - 2 ) = 7 . 1  \(\Rightarrow\hept{\begin{cases}y=6\\x=3\end{cases}}\)

TH2 ; ( y + 1 ) . ( x - 2  ) = 1 . 7 \(\Rightarrow\hept{\begin{cases}y=0\\x=9\end{cases}}\)

TH3 ; ( y + 1 ) . ( x - 2 ) = ( - 1 ) . ( - 7 ) \(\Rightarrow\hept{\begin{cases}y=-2\\x=-5\end{cases}}\)

TH4 : ( y + 1 ) . ( x - 2 ) = ( - 7 ) . ( - 1 )  \(\Rightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)