(x+2).(y-3)=-3=-1.3=1.(-3)

Vì x,y thuộc Z nên ( x+2) và (y+3) thuộc Z

Ta có bảng:

Vậy nếu x = - 3 thì y = 0

       nếu x = -1 thì y =- 6

       nếu x = - 5 thì y = - 2

       nếu x = 1 thì y = - 4

sao ko làm hết  bài

a) 1 - 2 + 3 - 4 + 5 - 6 + ... + 159 - 160

= (1 - 2) + (3 - 4) + (5 - 6) + ... + (159 - 160)

= (-1) + (-1) + (-1) + ... + (-1)

       có 80 số -1

= (-1) . 80

= -80

1 - 2 + 3 - 4 + 5 - 6 + ......... + 159 - 160 ( có 160 số )

= - 1 + ( - 1 ) + ( - 1 ) + .......... + ( - 1 )    ( có 80 số - 1 )

= - 1 . 80

= - 80 

1.

a,  \(x-14=3x+18\)                                                                       

\(\Rightarrow x-3x=18+14\)                                                                 

\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)

b, \(\left(x+7\right).\left(x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)

c, \(\left|2x-5\right|-7=22\)                                                                     

\(\Rightarrow\left|2x-5\right|=22+7\)

\(\Rightarrow\left|2x-5\right|=29\)

\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)

d,  \(\left(\left|2x\right|-5\right)-7=22\)

\(\Rightarrow\left(\left|2x\right|-5\right)=29\)

\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)

e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)

Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)

Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)

\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)

     \(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)

      \(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)

Ta có : 

\(x+3+x+9+x+5=4x\)

\(\Rightarrow3x+\left(3+9+5\right)=4x\)

\(\Rightarrow4x-3x=17\)

\(\Rightarrow x=17\)

2. a , b sai đề bn 

c, \(\left(5x+1\right).\left(y-1\right)=4\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)

\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Ta có bảng sau : 

d, \(5xy-5x+y=5\)

\(\Rightarrow\left(5xy-5x\right)+y=5\)

\(\Rightarrow5x.\left(y-1\right)+y=5\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)

\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Ta có bảng sau : 

x - 14 = 3x + 18

x - 3x = 18 + 14

-2x= 32

x= 32 : (-2)

x=-16

y=2x+3/x+1=2-(3/x+1) suy ra x+1 thuộc ước của 3 gồm 1;-1;3;-3

vay x=0 thi y=2.0+3/0+1=3

       x=-2 thi y=2.(-2)+3/(-2)+1=1

       x=2 thi y=2.2+3/3+1=7/4 vay x khong the bang 2

       x=-4 thì y=2.(-4)+3/-4+1=5/-3 vậy x không thể bằng -4

       vay x=0;-2 thi y=3;1

a) Ta có: (x-2)(y+1)=-1

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-1\\y+1=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)

Vậy: (x,y)={(3;-2);(1;0)}

b) Ta có: \(\left(2x+1\right)\left(y-2\right)=3\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+1=1\\y-2=3\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=3\\y-2=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=-1\\y-2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=-3\\y-2=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x=0\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}2x=2\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}2x=-2\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}2x=-4\\y=1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy: (x,y)={(0;5);(1;3);(-1;-1);(-2;1)}

\(\frac{x}{4}=\frac{y}{3}\)

\(\Rightarrow\frac{x+y}{4+3}=\frac{x}{4}=\frac{y}{3}\) mà x + y = 14

\(\Rightarrow\frac{14}{7}=\frac{x}{4}=\frac{y}{3}\)

\(\Rightarrow2=\frac{x}{4}=\frac{y}{3}\)

\(\Rightarrow\hept{\begin{cases}x=2\cdot4=8\\y=2\cdot3=6\end{cases}}\)