C,(x+5)(y-3)=15
D,xy+x+y=2
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(15xy^2z^3:3xyz^2=5yz\)
b: \(12x^4y^4:\left(-4x^4y^2\right)=-3y^2\)
c: \(\dfrac{-15x^2y^3z^2}{-6xz^2}=\dfrac{5}{2}xy^3\)
d: \(\dfrac{\left(x-y\right)^5}{\left(y-x\right)^3}=-\left(x-y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) Ta có:
2x=5y=>x/5=y/2
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
x/5=y/2=x-y/5-2=15/3=5
=> x=5.5=25; y=5.2=10
d)Đặt x/2=y/5=k
=> x=2k; y=5k=> xy=2k.5k=10k^2=10=> k^2=1=>k=\(\pm\)1
Với k=1=>x=2; y=5
Với k=-1=>x=-2; y=-5
![](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)
đối với các câu này bạn hãy khai triển phần nào dài bằng hàng dẳng thức rồi thu gọn lại nếu đúng thì vế trái bằng vế phải
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)
Thay x = 15 vào bt A ta có
A = 9 . 15 = 135
b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)
Thay x = -1/5 ; y = - 1/2 vào bt B ta có
\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(=9x^2y^2-xy^3-8x^3\)
Thay x = 1/2 ; y = 2 vào bt C ta có
\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)
d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)
\(=12x^2+12x-3\)
\(\left|x\right|=2\Rightarrow x=\pm2\)
Thay x = 2 vào bt D có
\(D=12.4+12.2-3=69\)
Thay x = - 2 vào bt D ta có
\(D=12.4-12.2-3=21\)
![](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.
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