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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
(x+1)(y-2) = 3
=> x+1 và y-2 là các ước của 3
Ư(3) = {1; -1; 3; -3}
Lập bảng giá trị:
x+1 | 1 | 3 | -1 | -3 |
y-2 | 3 | 1 | -3 | -1 |
x | 0 | 2 | -2 | -4 |
y | 5 | 3 | -1 | 1 |
Vậy các cặp (x,y) cần tìm là:
(0; 5); (2; 3); (-2; -1); (-4; 1).
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-2\right)\left(y+3\right)=15\)
\(\Rightarrow\left(x-2\right)\left(y+3\right)=1.15=15.1=\left(-1\right).\left(-15\right)=\left(-15\right).\left(-1\right)=3.5=5.3=\left(-3\right).\left(-5\right)=\left(-5\right).\left(-3\right)\)
Ta có bảng sau:
\(x-2\) | \(1\) | \(15\) | \(-1\) | \(-15\) | \(3\) | \(5\) | \(-3\) | \(-5\) |
\(y+3\) | \(15\) | \(1\) | \(-15\) | \(-1\) | \(5\) | \(3\) | \(-5\) | \(-3\) |
\(x\) | \(3\) | \(17\) | \(1\) | \(-13\) | \(5\) | \(7\) | \(-1\) | \(-3\) |
\(y\) | \(12\) | \(-2\) | \(-18\) | \(-4\) | \(2\) | \(0\) | \(-8\) | \(-6\) |
KL: Các cặp số (x; y)...
b) \(\left(3x+2\right)\left(1-y\right)=-7\)
\(\Rightarrow\left(3x+2\right)\left(1-y\right)=1.\left(-7\right)=\left(-7\right).1=\left(-1\right).7=7.\left(-1\right)\)
Ta có bảng sau:
\(3x+2\) | \(1\) | \(-7\) | \(-1\) | \(7\) |
\(1-y\) | \(-7\) | \(1\) | \(7\) | \(-1\) |
\(x\) | \(-\dfrac{1}{3}\) | \(-3\) | \(-1\) | \(\dfrac{5}{3}\) |
\(y\) | \(8\) | \(0\) | \(-6\) | \(2\) |
KL: Các cặp số (x; y)...
c) \(xy-5x=14-\left(-1\right)\)
\(\Leftrightarrow x\left(y-5\right)=15\)
\(\Rightarrow x\left(y-5\right)=1.15=15.1=\left(-1\right).\left(-15\right)=\left(-15\right).\left(-1\right)=3.5=5.3=\left(-3\right).\left(-5\right)=\left(-5\right).\left(-3\right)\)
Ta có bảng sau:
\(x\) | \(1\) | \(15\) | \(-1\) | \(-15\) | \(3\) | \(5\) | \(-3\) | \(-5\) |
\(y-5\) | \(15\) | \(1\) | \(-15\) | \(-1\) | \(5\) | \(3\) | \(-5\) | \(-3\) |
\(y\) | \(20\) | \(6\) | \(-10\) | \(4\) | \(10\) | \(8\) | \(0\) | \(2\) |
KL: Các cặp số (x; y)...
c') \(xy+x=5\)
\(\Leftrightarrow x\left(y+1\right)=5\)
\(\Rightarrow x\left(y+1\right)=1.5=5.1=\left(-1\right).\left(-5\right)=\left(-5\right).\left(-1\right)\)
Ta có bảng sau:
\(x\) | \(1\) | \(5\) | \(-1\) | \(-5\) |
\(y+1\) | \(5\) | \(1\) | \(-5\) | \(-1\) |
\(y\) | \(4\) | \(0\) | \(-6\) | \(-2\) |
KL: Các cặp số (x; y)...
d) Chưa tìm ra cách giải, chờ đã...
![](https://rs.olm.vn/images/avt/0.png?1311)
a,Ta có:\(xy+x=3\)
\(\Leftrightarrow x\left(y+1\right)=3\)
Vì x,y thuộc Z \(\hept{\begin{cases}x\\y+1\end{cases}}\in Z\)
\(\Rightarrow x;y+1\inƯ\left(3\right)\)
\(\Rightarrow x;y+1\in\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\y+1=3\Rightarrow y=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\y+1=-3\Rightarrow y=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\y+1=1\Rightarrow y=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\y+1=-1\Rightarrow y=-2\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d) \(x.\left(y+2\right)-y=15\)
\(\Rightarrow x.\left(y+2\right)=15+y\)
\(\Rightarrow x=\frac{y+15}{y+2}=\frac{y+2+13}{y+2}=1+\frac{13}{y+2}\)
y + 2 là ước nguyên của 13
\(y+2=1\Rightarrow y=-1\Rightarrow x=14\)
\(y+2=-1\Rightarrow y=-3\Rightarrow x=-12\)
\(y+2=13\Rightarrow y=11\Rightarrow x=2\)
\(y+2=-13\Rightarrow y=-15\Rightarrow x=0\)
Ai thấy đúng thì ủng hộ, mink chỉ làm được vậy thuu
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này thêm điều kiện là: x,y thuộc Z nha ko là ko lm đc đâu
a, (x+5)(y-3)=15
x+5 | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
y-3 | -1 | -3 | -5 | -15 | 15 | 5 | 3 | 1 |
x | -20 | -10 | -8 | -6 | -4 | -2 | 0 | 10 |
y | 2 | 0 | -2 | -12 | 18 | 8 | 6 | 4 |
Vậy có 8 cặp(x;y):...
các ý còn lại tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài giải
Mình làm câu a các câu b , d bạn làm tương tự nha !
a, \(\left(x+5\right)\left(y-3\right)=15\)
\(\Rightarrow\text{ }x+5\text{ , }y-3\inƯ\left(15\right)\)
x + 5 | - 1 | 1 | - 3 | 3 | - 5 | 5 | - 15 | 15 |
y - 3 | - 15 | 15 | - 5 | 5 | - 3 | 3 | - 1 | 1 |
x | - 6 | - 4 | - 8 | - 2 | - 10 | 0 | - 20 | 10 |
y | - 12 | 18 | - 2 | 8 | 0 | 6 | - 2 | 4 |
Vậy các cặp \(\left(x,y\right)=\text{ }...\)
c, \(xy+y+x=30\)
\(y\left(x+1\right)+x=30\)
\(y\left(x+1\right)+\left(x+1\right)=31\)
\(\left(y+1\right)\left(x+1\right)=31\)
Đến đây làm tương tự câu a nha !
Câu e để mình nghĩ tí đã nha !
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 200 - x = (-16)
x = 200 - (-16)
x = 200 + 16
x = 216
Vậy x = 216
b) 2x - 17 = 33
2x = 33 + 17
2x = 50
x = 50 : 2
x = 25
Vậy x = 25
c) \(\left|x\right|\)+ 6 = 15
\(\left|x\right|\) = 15 - 6
\(\left|x\right|\) = 9
=> x = 9 hoặc x = -9
d) 35 - x = (-5)
x = 35 - (-5)
x = 35 + 5
x = 40
Vậy x = 40
e) \(\left|x+1\right|\) - 3 = 5
\(\left|x+1\right|\) = 5 + 3
\(\left|x+1\right|\) = 8
=> x + 1 = 8 hoặc x + 1 = -8
* x + 1 = 8
x = 8-1
x = 7
* x + 1 = (-8)
x = (-8)-1
x = -9
Vậy x = 7 hoặc x = -9
f) 3y - 250 = 200
3y = 200 + 250
3y = 450
y = 450 : 3
y = 150
Vậy y = 150
g) ( xin lỗi câu này mình chưa nghĩ ra được)
h) \(\left|x-1\right|\) = 15
=> x - 1 = 15 hoặc x - 1 = -15
* x - 1 = 15
x = 15 + 1
x = 16
* x - 1 = -15
x = (-15) + 1
x = -14
Vậy x= 16 hoặc x = -14
a, 200 - x = -16
=> x = 200 - (-16)
=> x = 216
b, 2x -17 = 33
=> 2x = 33 + 17
=> 2x = 50
=> x = 50 : 2
=> x = 25
c, |x| + 6 = 15
=> |x| = 15 - 6
=> |x| = 9
=> x = -9 ; 9
d, 35- x = -5
=> x = 35 - (-5)
=> x = 40
e, |x + 1| -3 = 5
=> | x+1| = 5 - (-3)
=> |x+1 | = 8
=> x + 1 = -8 ; 8
* x + 1 = -8
=> x = -8 - 1
=>x = -9
* x + 1 = 8
=> x = 8- 1
=> x = 7
f, 3y - 250 = 200
=> 3y = 200 + 250
=> 3y = 450
=> y = 450 : 3
=> x = 150
g, x ( y- 3) = 21
=> xy - 3x = 21
Nếu xy = 7 thì 3x = 3 vậy x = 1
Nếu xy = 3 thì 3x = 7 vậy x là tập hợp rỗng
Vậy x = 1
h, |x -1| = 15
=> x - 1 = -15; 15
* x- 1 = -15
=> x = -15 + 1
=> x = - 14
* x - 1 = 15
=> x = 15 + 1
=> x =16
Vậy x = -14 ; 16
i, 25 - 3 . | 7- y| =1
=> 3. | 7- y | = 25 - 1
=> 3. | 7-y| = 24
=> | 7-y| = 24 : 3
=> | 7-y| = 8
=> 7-y= -8 ; 8
* 7-y= -8
=> y = 7 - ( -8)
=> y = 15
* 7-y= 8
=> y = 7 - 8
=> y =-1
![](https://rs.olm.vn/images/avt/0.png?1311)
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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C, (x+5)(y-3)=15 phần này bạn chia trường hợp ra nhé
vd 15=3*5 =>x+5=3 và y-3=15 tương tự làm tiếp nhé
D, xy+x+y=2
=>xy+x+y+1=2+1
=>(x+1)(y+1)=3 làm tương tự nhé