1, XY = -11
2 , x(x-2)=0
3, (2x+1)y=-14
4, (2y-1)x=-18
giup minh nha
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b: \(C=xy\left(x^3+2\right)-y\left(xy^3+2x\right)\)
\(=x^4y+2xy-xy^4-2xy\)
\(=xy\left(x^3-y^3\right)\)
\(=xy\left(x-y\right)\left(x^2+xy+y^2\right)⋮x^2+xy+y^2\)
\(VT=\dfrac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}=\dfrac{\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x-y}\)
bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)
\(=\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x-x^2-2x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{x^2-4}\)
Bài 2:
1: \(x^2y^2-8-1\)
\(=x^2y^2-9\)
\(=\left(xy-3\right)\left(xy+3\right)\)
2: \(x^3y-2x^2y+xy-xy^3\)
\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)
\(=xy\left(x^2-2x+1-y^2\right)\)
\(=xy\left[\left(x-1\right)^2-y^2\right]\)
\(=xy\left(x-1-y\right)\left(x-1+y\right)\)
3: \(x^3-2x^2y+xy^2\)
\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)
\(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)
4: \(x^2+2x-y^2+1\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
5: \(x^2+2x-4y^2+1\)
\(=\left(x^2+2x+1\right)-4y^2\)
\(=\left(x+1\right)^2-4y^2\)
\(=\left(x+1-2y\right)\left(x+1+2y\right)\)
6: \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
m: (x-y)(x^2-2xy+y^2)
=(x-y)*(x-y)^2
=(x-y)^3
=x^3-3x^2y+3xy^2-y^3
n: =-(x^3+x^2y-x-x^2y-xy^2+y)
=-x^3+x+xy^2-y
o: =-(x^3+x^2y^2-x^2-2xy-2y^3+2y)
=-x^3-x^2y^2+x^2+2xy+2y^3-2y
p: (1/2x-1)(2x-3)
=1/2x*2x-1/2x*3-2x+3
=x^2-3/2x-2x+3
=x^2-7/2x+3
q: (x-1/2y)(x-1/2y)
=(x-1/2y)^2
=x^2-xy+1/4y^2
r: (x^2-2x+3)(1/2x-5)
=1/2x^3-5x^2-x^2+10x+3/2x-15
=1/2x^3-6x^2+11,5x-15
a) \(x\left(x^2-2x\right)+\left(x-2x\right)=x^2\left(x-2\right)+x\left(x-2\right)=\left(x-2\right)\left(x^2+x\right)⋮x-2\forall x,y\in Z\)
b) \(x^3y^2-3yx^2+xy=xy\left(x^2y-3x+1\right)⋮xy\forall x,y\in Z\)
c) \(x^3y^2-3x^2y^3+xy^2=xy^2\left(x^2-3xy+1\right)⋮\left(x^2-3xy+1\right)\forall x,y\in Z\)
1.
x.y=-11
=> x,y thuộc Ư(-11)={-1,-11,1,11}
Ta có bảng :
Vậy ta có các cặp x,y thõa mãn là : (-1,-11);(-11,-1);(1,11);(11,1)
2) x(x-2)=0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy x=0 hoặc x=2
3) (2x+1)y=-14
=> 2x+1 ; y thuộc Ư(-14)={-1,-2,-7,-14,1,2,7,14}
Ta có bảng :
Vậy ...
4) (2y-1)x=-18
=> 2y-1 ; x thuộc Ư(-18)={-1,-2,-3,-6,-9,-18,1,2,3,6,9,18}
Ta có bảng :
Vậy ...