A=1/3x^2y-1/3x^2y+xy^2-xy+1/2xy^2-5xy

=3/2xy^2-6xy

=3/2*1/2*1^2-6*1/2*1

=3/4-3=-9/4

`@` `\text {Ans}`

`\downarrow`

`A = 1/3x^2y + xy^2 - xy + 1/2xy^2 - 5xy - 1/3x^2y`

`= (1/3 x^2y - 1/3x^2y) + (xy^2 + 1/2xy^2) + (-xy - 5xy)`

`= 3/2 xy^2 - 6xy`

Thay `x = 1/2; y = 1` vào A

`A = 3/2* 1/2 * 1^2 - 6*1/2 * 1`

`= 3/4 - 3`

`= -9/4`

Vậy, `A = -9/4.`

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

a/ \(C=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-1\right)\)

\(C=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-1\right)=x+y-1\) (do x+y-2=0)

Mà x+y-2=0 => x+y-1=1 => C=1

b/  Với x=2; y=2 Ta nhận thấy \(x^3-2y^2=2^3-2.2^2=2^3-2^3=0\) => D=0

Đặt \(A=\left(\dfrac{2}{5}x^3y\right)\cdot\left(-5xy\right)\)

\(=\left(\dfrac{2}{5}\cdot\left(-5\right)\right)\cdot x^3\cdot x\cdot y\cdot y\)

\(=-2x^4y^2\)

Thay x=-1 và y=1/2 vào A, ta được:

\(A=-2\cdot\left(-1\right)^4\cdot\left(\dfrac{1}{2}\right)^2=-2\cdot\dfrac{1}{4}=-\dfrac{1}{2}\)

\(=x^4-xy+xy+x^2y-x^4-x^2y+3xy-xy.\)

\(=2xy\)

Thay  x = 1/4 , y = - 2005 ta được: 2xy = 2.1/4 .  ( - 2005 ) = -2005/2