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a) \(\left(x-1\right)\left(y+2\right)=7\)
\(\Leftrightarrow\left(x-1\right)\left(y+2\right)=1.7=7.1=\left(-1\right).\left(-7\right)=\left(-7\right).\left(-1\right)\)
Ta có bảng sau:
| \(x-1\) | \(1\) | \(7\) | \(-1\) | \(-7\) |
| \(y+2\) | \(7\) | \(1\) | \(-7\) | \(-1\) |
| \(x\) | \(2\) | \(8\) | \(0\) | \(-6\) |
| \(y\) | \(5\) | \(-1\) | \(-9\) | \(-3\) |
Vậy \(\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\left\{{}\begin{matrix}x=8\\y=-1\end{matrix}\right.\left\{{}\begin{matrix}x=0\\y=-9\end{matrix}\right.\left\{{}\begin{matrix}x=-6\\y=-3\end{matrix}\right.\)
b) \(\left(x-2\right)\left(2y+1\right)=17\)
\(\Rightarrow\left(x-2\right)\left(2y+1\right)=1.17=17.1=\left(-1\right).\left(-17\right)=\left(-17\right).\left(-1\right)\)
Ta có bảng sau:
| \(x-2\) | \(1\) | \(17\) | \(-1\) | \(-17\) |
| \(2y+1\) | \(17\) | \(1\) | \(-17\) | \(-1\) |
| \(x\) | \(3\) | \(19\) | \(1\) | \(-15\) |
| \(y\) | \(8\) | \(0\) | \(-9\) | \(-1\) |
Vậy \(\left\{{}\begin{matrix}x=3\\y=8\end{matrix}\right.\left\{{}\begin{matrix}x=19\\y=0\end{matrix}\right.\left\{{}\begin{matrix}x=1\\y=-9\end{matrix}\right.\left\{{}\begin{matrix}x=-15\\y=-1\end{matrix}\right.\)
a) ( x - 1 ) . ( y + 2 ) = 7
Lập bảng ta có :
| x-1 | 1 | 7 | -1 | -7 |
| y+2 | 7 | 1 | -7 | -1 |
| x | 2 | 8 | 0 | -6 |
| y | 5 | -1 | -8 | -3 |
b) x . ( y - 3 ) = -12
Lập bảng ta có :
| y-3 | 12 | -12 | 2 | -2 | -3 | -4 |
| x | -1 | 1 | -6 | 6 | 4 | 3 |
| y | 15 | -9 | 5 | 1 | 0 | -1 |
c) xy - 3x - y = 0
x . ( y - 3 ) - y = 0
x . ( y - 3 ) - y + 3 = 3
x . ( y - 3 ) - ( y - 3 ) = 3
( x - 1 ) . ( y - 3 ) = 3
Lập bảng ta có :
| x-1 | 3 | 1 | -1 | -3 |
| y-3 | 1 | 3 | -3 | -1 |
| x | 4 | 2 | 0 | -2 |
| y | 4 | 6 | 0 | 2 |
d) xy + 2x + 2y = -16
x . ( y + 2 ) + 2y = -16
x . ( y + 2 ) + 2y + 4 = -12
x . ( y + 2 ) + 2 . ( y + 2 ) = -12
( x + 2 ) . ( y + 2 ) = -12
Lập bảng ta có :
| x+2 | 1 | -1 | -2 | -6 | -4 | -3 |
| y+2 | -12 | 12 | 6 | 2 | 3 | 4 |
| x | -1 | -3 | -4 | -8 | -6 | -5 |
| y | -14 | 10 | 4 | 0 | 1 | 2 |
Ta có : (x - 1).(y + 2) = 7
=> (x - 1) và y + 2 thuộc Ư(7) = {-7;-1;1;7}
Ta có bảng :
| x - 1 | -7 | -1 | 1 | 7 |
| y + 2 | -1 | -7 | 7 | 1 |
| x | -6 | 0 | 2 | 8 |
| y | -3 | -9 | 5 | -1 |
Vậy có 4 cặp x;y thoả mãn : (-6,-3) ; (0 , -9) ; (2 , 5) ; (8, -1)
a)x.y=-2
\(\Rightarrow\)\(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)hoặc \(\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
b) mik lỡ bấm nhầm câu hỏi kề câu hỏi của bạn
Đây nek : https://olm.vn/hoi-dap/detail/238833793861.html
a.
xy + 3x - 2y - 6 = 5
=>x(y + 3) - 2(y + 3) = 5
=>(x - 2)(y + 3) = 5.
Vì x, y thuộc Z nên x - 2, y + 3 thuộc Z
=> x - 2, y + 3 thuộc ước nguyên của 5
Lập bảng :
| x - 2 | -5 | -1 | 1 | 5 |
| y + 3 | -1 | -5 | 5 | 1 |
| x | -3 | 1 | 3 | 7 |
| y | -4 | -8 | 2 | -2 |
Vậy ......
b. Làm tương tự câu a.
c. Ta có x + y = 3 và x - y = 15
Bài này là tổng hiệu của cấp 1, áp dụng cách làm đó thì ta được số lớn là x = (3 + 15) : 2 = 9
Số bé là y = 9 - 15 = -6
d. Ta có : |x| + |y| = 1
=>|x| = 1 - |y|
Vì |x|, |y| >= 0 và |x| = 1 - |y| nên 0 =< |x|, |y| =< 1
Vì x, y thuộc Z nên x = 0 thì y = 1 hoặc -1 và ngược lại y = 0 thì x = 1 hoặc -1
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Bài 1: Ta có 5x+7=5(x-2)+8
Để 5x+7 chia hết cho x-2 thì 5(x-2) +8 chia hết cho x-2
=> 8 chia hết cho x-2
x nguyên => x-2 nguyên => x-2 thuộc Ư (8)={-8;-4;-2;-1;1;2;4;8}
ta có bảng
| x-2 | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| x | -6 | -2 | 0 | 1 | 3 | 4 | 6 | 10 |
Bài 2:
a) xy+x=-15
<=> x(y+1)=-15
=> x, y+1 thuộc Ư (-15)={-15;-5;-3;-1;1;3;5;15}
Ta có bảng
| x | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
| y+1 | 1 | 3 | 5 | 15 | -15 | -5 | -3 | -1 |
| y | 0 | 2 | 4 | 14 | -16 | -6 | -4 | -2 |
b) xy+2-y=9
<=> y(x-1)=7
=> y, x-1 thuộc Ư (7)={-7;-1;1;7}
Ta có bảng
| y | -7 | -1 | 1 | 7 |
| x-1 | -1 | -7 | 7 | 1 |
| x | 0 | -6 | 6 | 2 |
c) xy+2x+2y=-17
<=> x(y+2)+2(y+2)=-15
<=> (x+2)(y+2)=-15
<=> x+2; y+2 thuộc Ư (-15)={-15;-5;-3;-1;1;3;5;15}
Ta có bảng
| x+2 | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
| x | -17 | -7 | -5 | -3 | -1 | 1 | 3 | 13 |
| y+2 | 1 | 3 | 5 | 15 | -15 | -5 | -3 | -1 |
| y | -1 | 1 | 3 | 13 | -17 | -7 | -5 | -3 |
a) xy + x + 2y = 5
=> (xy + x) + 2y + 2 = 7
=> x(y + 1) + 2(y + 1) = 7
=> (x + 2)(y + 1) = 7
=. x + 2 \(\in\) (7) = {-1; -7; 1; 7}
Ta có bảng sau:
| x + 2 | -1 | 1 | -7 | 7 |
| x | -3 | -1 | -9 | 5 |
| y + 1 | -7 | 7 | -1 | 1 |
| y | -8 | 6 | -2 | 0 |
Vậy (x; y) \(\in\){(-3; -8); (-1; 6); (-9; -2); (5; 0)}
a) <=> (xy+x) + 2y + 2 = 7
=> x(y+1) + 2(y+1) = 7
=> (x+2)(y+1) = 7
Vì x nguyên => x+2 \(\in\) Ư(7)= {7;-7;1;-1}
Ta có bảng sau:
| x+2 | 7 | -7 | 1 | -1 |
| x | 5 | -9 | -1 | -3 |
| y+1 | 1 | -1 | 7 | -7 |
| y | 0 | -2 | 6 | -8 |
Vậy (x;y) = (5;0); (-9;-2) ; (-1;6); (-3;-8)
a/ \(\left(x-1\right)\left(y+2\right)=7\)
\(\Rightarrow\left(x-1\right);\left(y+2\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
TH1: x-1=1
x=2
y+2=7
y=5
TH2: x-1=-1
x=0
y+2=-7
y=-9
TH3: x-1=7
x=8
y+2=1
y=-1
TH4: x-1=-7
x=-6
y+2=-1
y=-3
b/\(\left(x-2\right)\left(2y+1\right)=17\)
\(\Rightarrow\left(x-2\right);\left(2y+1\right)\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
TH1: x-2=1
x=3
2y+1=17
y=8
TH2: x-2=17
x=19
2y+1=1
y=0
TH3: x-2=-1
x=1
2y+1=-17
y=-9
TH4: x-2=-17
x=-15
2y+1=-1
y=-1
a) ( x - 1 ) . ( y + 2 ) = 7
<=> ( x - 1 ) ; ( y + 2 ) \(\in\) Ư(7) = { -7 ; -1 ; 1 ; 7 }
Vậy các cặp ( x ; y ) thỏa mãn yêu cầu đề bài là :
( - 6 ; - 3 ) ; ( 0 ; - 9 ) ; ( 2 ; 5 ) ; ( 8 ; - 1 )
b) ( x - 2 ) . ( 2y + 1 ) = 17
<=> ( x - 2 ) ; ( 2y + 1 ) \(\in\) Ư(17) = { - 17 ; -1 ; 1 ; 17 }
Ta có bảng sau :
Vậy các cặp ( x ; y ) thỏa mãn yêu cầu đề bài là :
( - 15 ; - 1 ) ; ( 1 ; -9 ) ; ( 3 ; 8 ) ; ( 19 ; 0 )
Câu c và câu d mk ko biết làm , xin lỗi bạn nha