\(a,-x^2y-2xy+2x^2y+5xy+2\\ =x^2y+3xy+2\\ b,-2xy+\dfrac{3}{2}xy^2+\dfrac{1}{2}xy^2+xy\\ =-xy+2xy^2\)

chọn D vì  \(\dfrac{7x-4}{2xy}+\dfrac{4-7x}{2xy}=0\)

`x^2 -4x+4-y^2`

`=(x^2 -4x+4)-y^2`

`=(x-2)^2 -y^2`

`=(x-2-y)(x-2+y)`

`x^2+2xy+y^2-x-y`

`=(x^2+2xy+y^2) -(x+y)`

`=(x+y)^2 -(x+y)`

`=(x+y)(x+y-1)`

`x^2-2xy+y^2-9`

`=(x^2-2xy+y^2)-3^2`

`=(x-y)^2-3^3`

`=(x-y-3)(x-y+3)`

Tách ra đi cậu.

https://hoc24.vn/cau-hoi/927x118.8505894378996

giúp mik với ạ

\(=4x^2y^2+2xy-6-4x^2y^2+4xy-1=6xy-7=6\cdot\dfrac{1}{3}\cdot3-7=6-7=-1\)

0,2:x=1,03+3,97

a: A=-2xy+xy+xy^2=-xy+xy^2

Bậc là 3

b: \(B=xy^2z+2xy^2z-3xy^2z+xy^2z-xyz=-xyz+xy^2z\)

Bậc là 4

c: \(C=4x^2y^3-x^2y^3+x^4+6x^4-2x^2=3x^2y^3+7x^4-2x^2\)

Bậc là 5

d: \(D=\dfrac{3}{4}xy^2-\dfrac{1}{2}xy^2+xy=\dfrac{1}{4}xy^2+xy\)

bậc là 3

e: \(E=2x^2-4x^2+3z^4-z^4-3y^3+2y^3\)

=-2x^2+2z^4-y^3

Bậc là 4

f: \(=3xy^2z+xy^2z+2xy^2z-4xyz=6xy^2z-4xyz\)

Bậc là 4

a.

(x^2 + y^2 - 2xy) + (x^2 + y^2 + 2xy)

= x^2 + y^2 - 2xy + x^2 + y^2 + 2xy

= (x^2 + x^2) + (y^2 + y^2) + (2xy - 2xy)

= 2x^2 + 2y^2

b.

(x^2 + y^2 - 2xy) - (x^2 + y^2 + 2xy)

= x^2 + y^2 - 2xy - x^2 - y^2 - 2xy

= (x^2 - x^2) + (y^2 - y^2) - (2xy + 2xy)

= -4xy

Sửa đề: 2x-y+2xy-7=0

Ta có: 2x-y+2xy-7=0

=>2xy+2x-y-7=0

=>2x(y+1)-y-1-6=0

=>2x(y+1)-(y+1)=6

=>(2x-1)(y+1)=6

=>(2x-1;y+1)∈{(1;6);(6;1);(-1;-6);(-6;-1);(2;3);(3;2);(-2;-3);(-3;-2)}

=>(2x;y)∈{(2;5);(7;0);(0;-7);(-5;-2);(3;2);(4;1);(-1;-4);(-2;-3)}

=>(x;y)∈{(1;5);(7/2;0);(0;-7);(-5/2;-2);(3/2;2);(2;1):(-1/2;-4);(-1;-3)}

\(\left(x+y-1\right)\left(x+y+1\right)=x^2+xy-x+xy+y^2-y+x+y-1\\ =x^2+\left(xy+xy\right)+\left(-x+x\right)+y^2+\left(-y+y\right)-1\\ =x^2+2xy+y^2-1\\ =>B\)

Ta có: \(\left(3x^3-2xyz+\frac12xy^2\right)\cdot\left(-2xy\right)\)

\(=-3x^3\cdot2xy+2xyz\cdot2xy-\frac12xy^2\cdot2xy\)

\(=-6x^4y+4x^2y^2z-x^2y^3\)