xy-7x+y=-22
xy-3x+y=-20
xy-5y-2x=-41
TÌM X
GIÚP MK NHA
(x+1).(xy-2)=11
xy-7x+y=-22
xy-3x+y=-20
xy-5y-2x=-41
lam ca cach trinh bay giup mk nhe
mk se tich dung cho
1. xy - 7x + y = -22
2. xy - 3x + y = -20
3. xy - 5y - 2x = -41
1. x[y-7]+[y-7]=-22-7
x[y-7]+1[y-7]=-29
[x+1][y-7]=-29
suy ra -29=-1.29=-29.1
tiếp theo tự làm
2.x[y-3]+[y-3]=-20-3
x[y-3]+1[y-3]=-23
[x+1][y-3] = -23
-23=-1. 23= -23.1
tiếp theo tự làm nhé
3. y[x-5]-2x-10=-41
y[x-5]-2[x-5]=-41
[y-2][x-5]=-41
-41=-1. 41=-41.1
tiếp theo tự làm
NHỚ TICK VÀ THEO DÕI MÌNH NHÉ
(x-3).(y+5)=-17
(x+1).(xy-2)=11
xy-7x+y=-22
xy-3x+y =-20
xy-5y-2x=-41
a)(x-3).(y+5)=-17
\(\Rightarrow-17⋮x-3\)
\(\Rightarrow x-3\inƯ\left(-17\right)=\left\{\pm1;\pm17\right\}\)
+)Ta có bảng:
x-3 | -1 | 1 | -17 | 17 |
y+5 | -17 | 17 | -1 | 1 |
x | 2\(\in Z\) | 4\(\in Z\) | -14\(\in Z\) | 20\(\in Z\) |
y | -22\(\in Z\) | 12\(\in Z\) | -6\(\in Z\) | -4\(\in Z\) |
Vậy \(\left(x,y\right)\in\left\{\left(2;-22\right);\left(4;12\right);\left(-14;-6\right);\left(20;-4\right)\right\}\)
Chúc bn học tốt
\(c,xy-7x+y=-22\)
\(=>x.\left(y-7\right)+y-7=-29\)
\(=>\left(x+1\right)\left(y-7\right)=-29\)
\(=>x+1;y-7\inƯ\left(-29\right)\)
Nên ta có bảng sau :
...
\(d,xy-3x+y=-21\)
\(=>x.\left(y-3\right)+y-3=-24\)
\(=>\left(x+1\right)\left(y-3\right)=-24\)
\(=>x+1;y-3\inƯ\left(-24\right)\)
Nên ta có bảng sau :
...
\(e,xy-5y-2x=-41\)
\(=>y.\left(x-5\right)-2.\left(x-5\right)=-31\)
\(=>\left(y-2\right).\left(x-5\right)=-31\)
Vì \(y-2;x-5\inƯ\left(-31\right)\)
Nên ta có bảng sau :
P/s : mấy cái này bạn tự lập bảng nhé :P
Tìm số nguyên x, biết
a, xy - 7x + y= -22
b, xy - 3x + y= -20
c, xy - 5y - 2x = -41
a,\(xy-7y+y=-22\)
\(=xy-7x+y-7+7=-22\)
\(=\left(xy-7x\right)+\left(y-7\right)=-29\)
\(=x\left(y-7\right)+\left(y-7\right)=-29\)
\(=\left(y-7\right)\left(x+1\right)=-29\)
Vì \(x,y\varepsilon Z\)nên\(\left(y-7\right),\left(x+1\right)\varepsilon Z\)
\(\Rightarrow\left(y-7\right);\left(x+1\right)\varepsilon B\left(-29\right)\)
Mà \(-29=-1.29=1.\left(-29\right)\)
Ta có 4TH :\(1,\hept{\begin{cases}y-7=-1\\x+1=29\end{cases}}\Rightarrow\hept{\begin{cases}y=6\\x=28\end{cases}}\left(TM\right)\)
\(2,\hept{\begin{cases}y-7=1\\x+1=-29\end{cases}\Rightarrow\hept{\begin{cases}y=8\\x=-30\end{cases}}}\)
\(3,\hept{\begin{cases}y-7=29\\x+1=-1\end{cases}\Rightarrow\hept{\begin{cases}y=36\\x=-2\end{cases}}}\)
\(4,\hept{\begin{cases}y-7=-29\\x+1=1\end{cases}\Rightarrow\hept{\begin{cases}y=-22\\x=0\end{cases}}}\)
Vậy có 4 cặp (x, y): \(\left(6;28\right);\left(8;-30\right);\left(36;-2\right);\left(-22;0\right)\)
Vì dài quá nên mk chỉ làm từng này thôi nhé, nếu mk đúng nha!
b, xy - 3x + y= -20
=> x(y - 3) + (y - 3) = -23
=> (x + 1)(y - 3) = -23
ta có bảng :
x+1 | -1 | 1 | -23 | 23 |
y-3 | 23 | -23 | 1 | - |
x | -2 | 0 | -24 | 22 |
y | 26 | -20 | 4 | 2 |
c, xy - 5y - 2x = -41
=> y(x - 5) - 2x + 10 = -31
=> y(x - 5) - 2(x - 5) = -31
=> (y - 2)(x - 5) = -31
y-2 | -1 | 1 | -31 | 31 |
x-5 | 31 | -31 | 1 | -1 |
y | 1 | 3 | -29 | 33 |
x | 36 | -26 | 6 | 4 |
tìm x,y biết
3/ xy - 7x + y = -22
4/ xy - 3x + y = -20
5/ xy – 5y – 2x = -41
3, x(y-7)+y-7=-29
(x+1)(y-7)=-29 => (x+1)\(\in\) Ư(29) (y-7)\(\in\) Ư(29)
lập bảng:
...
(x,y)=(-2,36),(28,6),(-30,8),(0,-22)
4,5, tương tự
mk k chắc là làm đúng đâu
Tìm số nguyn x; y:
1, ( x+1).(xy-2) =11
2,xy-7x+y = -22
3,xy-3x+y =-20
4,xy-5y-2x =-41
phân tích đa thức thành nhân tử
\(a)3x^3+6x^2y \)
\(b)2x^3-6x^2\)
\(c)18x^2-20xy\)
\(d)xy+y^2-x-y \)
\(e)(x^2y^2-8)^2-1\)
\(f)x^2-7x-8\)
\(g)10x^2(2x-y)+6xy(y-2x)\)
\(h)x^2-2x+1-y^2\)
\(i)2x(x+2)+x^2(-x-2)\)
\(k)-9+6x-x^2\)
\(l)8xy-2x^2-8y^2\)
\(m)3x^2+5x-3y^2-5y\)
a) 3x³ + 6x²y
= 3x².(x + 2y)
b) 2x³ - 6x²
= 2x².(x - 2)
c) 18x² - 20xy
= 2x.(9x - 10y)
d) xy + y² - x - y
= (xy + y²) - (x + y)
= y(x + y) - (x + y)
= (x + y)(y - 1)
e) (x²y² - 8)² - 1
= (x²y² - 8 - 1)(x²y² - 8 + 1)
= (x²y² - 9)(x²y² - 7)
= (xy - 3)(xy + 3)(x²y² - 7)
f) x² - 7x - 8
= x² - 8x + x - 8
= (x² - 8x) + (x - 8)
= x(x - 8) + (x - 8)
= (x - 8)(x + 1)
a: \(3x^3+6x^2y\)
\(=3x^2\cdot x+3x^2\cdot2y=3x^2\left(x+2y\right)\)
b: \(2x^3-6x^2=2x^2\cdot x-2x^2\cdot3=2x^2\left(x-3\right)\)
c: \(18x^2-20xy=2x\cdot9x-2x\cdot10y=2x\left(9x-10y\right)\)
d: \(xy+y^2-x-y\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(y-1\right)\)
e: \(\left(x^2y^2-8\right)^2-1\)
\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left(x^2y^2-7\right)\left(x^2y^2-9\right)\)
\(=\left(x^2y^2-7\right)\left(xy-3\right)\left(xy+3\right)\)
f: \(x^2-7x-8\)
\(=x^2-8x+x-8\)
\(=x\left(x-8\right)+\left(x-8\right)=\left(x-8\right)\left(x+1\right)\)
g: \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)
\(=2x\cdot\left(2x-y\right)\cdot5x-2x\cdot\left(2x-y\right)\cdot3y\)
\(=2x\left(2x-y\right)\left(5x-3y\right)\)
h: \(x^2-2x+1-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-1-y\right)\left(x-1+y\right)\)
i: \(2x\left(x+2\right)+x^2\left(-x-2\right)\)
\(=2x\left(x+2\right)-x^2\left(x+2\right)\)
\(=\left(x+2\right)\left(2x-x^2\right)=x\cdot\left(x+2\right)\left(2-x\right)\)
k: \(-x^2+6x-9=-\left(x^2-6x+9\right)\)
\(=-\left(x^2-2\cdot x\cdot3+3^2\right)=-\left(x-3\right)^2\)
l: \(-2x^2+8xy-8y^2\)
\(=-2\left(x^2-4xy+4y^2\right)\)
\(=-2\left(x-2y\right)^2\)
m: \(3x^2+5x-3y^2-5y\)
\(=3\left(x^2-y^2\right)+5\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+3y+5\right)\)
g) 10x²(2x - y) + 6xy(y - 2x)
= 10x²(2x - y) - 6xy(2x - y)
= 2x(2x - y)(5x - 3y)
h) x² - 2x + 1 - y²
= (x² - 2x + 1) - y²
= (x - 1)² - y²
= (x - y - 1)(x + y - 1)
i) 2x(x + 2) + x² (-x - 2)
= 2x(x + 2) - x²(x + 2)
= x(x + 2)(2 - x)
k) -9 + 6x - x²
= -(x² - 6x + 9)
= -(x - 3)²
l) 8xy - 2x² - 8y²
= -2(x² - 4xy + 4y²)
= -2(x - 2y)²
m) 3x² + 5x - 3y² - 5y
= (3x² - 3y²) + (5x - 5y)
= 3(x² - y²) + 5(x - y)
= 3(x - y)(x + y) + 5(x - y)
= (x - y)[3(x + y) + 5]
= (x - y)(3x + 3y + 5)
tìm x y
x^3-3x^2+7x-21=2y
xy-2x-3y=5
2xy-3x+5y=8
giups minhf vs
Rút gọn biểu thức:
\(\left(3x-2y\right)^3-\left(4x-5y\right)\left(16x^2+20xy+25y^2\right)+\left(y+2x\right)^3\)
\(\left(3x-2y\right)^3+\left(y+2x\right)^3-\left(4x-5y\right)\left(16x^2+20xy+25y^2\right)\)
\(=27x^3-54x^2y+36xy^2-8y^3+y^3+6xy^2+12x^2y+8x^3-\left(64x^3-125y^3\right)\)
\(=35x^3-42x^2y+42xy^2-7y^3-64x^3+125y^3\)
\(=-29x^3-42x^2y+42xy^2+118y^3\)