2xy-3+4x-y=0
tim x,y thuoc z
a) ( x+3 ) * ( x^2 - 3x +9 ) - ( 54+ x^3 )
b) ( 2x + y ) * ( 4x^2 - 2xy + y^2 ) - ( 2x - y ) * ( 4x^2 + 2xy + y^2 )
c) ( a+b ) ^3 - ( a-b ) ^3 - 2b^3
d) ( x+y+z ) ^ 2 - 2 * ( x+y+z ) * ( x+y ) + y^2 + ( x + y ) ^ 2
a) \(\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3+27-54-x^3\)
\(=-27\)
b) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(8x^3+y^3\right)-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
c) \(\left(a+b\right)^3-\left(a-b\right)^3-2b^3\)
\(=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)-\left(a-b\right)^2\right]-2b^3\)
\(=\left(a+b-a+b\right)\left[\left(a^2+2ab+b^2\right)+\left(a^2-ab+ab-b^2\right)-\left(a^2-2ab+b^2\right)\right]-2b^3\)
\(=b^2\left(a^2+2ab+b^2+a^2-ab+ab-b^2-a^2+2ab-b^2\right)-2b^3\)
....
tim x,y thuoc z biet: x - y = 6 - 2xy
Ta có x-y=6-2xy
<=> x-y-6+2xy=0
<=> 2x-2y-12+4xy=0
<=> 2x(1+2y)-(1+2y)=11
<=> (1+2y)(2x-1)=11
=> 1+2y và 2x-1 là ước của 11
Ta có bảng sau
2x-1 | -11 | -1 | 1 | 11 |
x | -5 | 0 | 1 | 6 |
1+2y | -1 | -11 | 11 | 1 |
y | -1 | -6 | 5 | 0 |
Vậy ....
tim x,y thuoc z biet: x - y = 6 - 2xy
Ta có :
x-y = 6- 2xy
x-y+2xy = 6
x(2y+1) - y = 6
2x(2y+1) - 2y = 12
2x(2y+1) - (2y+1) = 11
(2x-1)(2y+1) = 11
Suy ra 2x-1 và 2y+1 là ước của 11
Các ước của 11 là:1;-1;11;-11
Ta có bảng sau:
2y+1 | 1 | 11 | -1 | -11 |
2x-1 | 11 | 1 | -11 | -1 |
y | 0 | -1 | 5 | -6 |
x | 6 | -5 | 1 | 0 |
Vậy_________________
tìm x,y thuoc Z
2xy-y+x=2
Rút gọn biểu thức:
a) (x-2)^3-x(x+1)(x-1)+6x(x-3)
b) (2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)
c) (x+y+z)^2-2(x+y+z)(x+y)+(x+y)
giúp mình vs!!!!
\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)
\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)
\(=-5x-27\)
\(b,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y+z-x-y\right)^2\)
\(=z^2\)
a)
=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\)
=-5x-19
b)
=\(8x^3+y^3-8x^3+y^3\)
=\(2y^3\)
c)
=(x+y+z-x-y)\(^2\) +x+y
=\(z^2+x+y\)
hc tốt
Tim x , y thuoc z , biet :
2xy + x + y = 21
ta có 2xy + x + y = 21
=> 4xy + 2x + 2y = 42
=> (4xy + 2x) + 2y = 42
=> 2x(2y+1) + 2y + 1 = 43
=> (2x + 1)(2y+1) = 43
\(\Rightarrow\hept{\begin{cases}2x+1\in\left\{1;43;-43;-1\right\}\\2y+1\in\left\{43;1;-1;-43\right\}\end{cases}}\Rightarrow\hept{\begin{cases}x\in\left\{0;21;-22;-1\right\}\\y\in\left\{21;0;-1;-22\right\}\end{cases}}\)
tim x, y thuoc Z, biet:
x - y + 2xy bang 7
nhanh nhe! Minh chuan bi thi roi
Ai nhanh ma dung, minh se k cho!
\(x-y+2xy=7\)
\(\Rightarrow x\left(2y+1\right)-y=7\)
\(\Rightarrow x\left(2y+1\right)=7+y\)
\(\Rightarrow2x.\left(2y+1\right)=2\left(7+y\right)\)
\(\Rightarrow2x\left(2y+1\right)=14+2y\)
\(\Rightarrow2x\left(2y+1\right)-\left(2y+1\right)=\left(14+2y\right)-\left(2y+1\right)\)
\(\Rightarrow\left(2x-1\right)\left(2y+1\right)=13\)
\(TH1:\hept{\begin{cases}2x-1=-1\\2y+1=-13\end{cases}}\Rightarrow\hept{\begin{cases}2x=0\\2y=-14\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=-7\end{cases}}\)
\(TH2:\hept{\begin{cases}2x-1=-13\\2y+1=-1\end{cases}}\Rightarrow\hept{\begin{cases}2x=-12\\2y=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=-6\\y=-1\end{cases}}\)
\(TH3:\hept{\begin{cases}2x-1=1\\2y+1=13\end{cases}}\Rightarrow\hept{\begin{cases}2x=2\\2y=12\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=6\end{cases}}\)
\(TH4:\hept{\begin{cases}2x-1=13\\2y+1=1\end{cases}}\Rightarrow\hept{\begin{cases}2x=14\\2y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\y=0\end{cases}}\)
Vậy các cặp giá trị \(\left(x;y\right)\)thoả mãn là: \(\left(0;-7\right)\), \(\left(-6;-1\right)\), \(\left(1;6\right)\), \(\left(7;0\right)\)
tim x,y thuoc Z biết 8y-x=2xy
Tim cap x,y biet x,y thuoc Z va
2xy- 4+2x+y=0