\(P=\dfrac{1}{3}x^2y+xy^2-xy+\dfrac{1}{2}xy^2-5xy-\dfrac{1}{3}x^2y=\dfrac{3}{2}xy^2-6xy\)

Thay x = 2 ; y = 1 ta được 

\(\dfrac{3}{2}.2.1-6.2.1=3-12=-9\)

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

=3/2xy^2-6xy

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

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

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

Bài giải:

\(\Leftrightarrow P=\left(\frac{1}{3}x^2y-\frac{1}{3}x^2y\right)+\left(xy^2+\frac{1}{2}xy^2\right)-\left(xy+5xy\right)\)

\(\Leftrightarrow P=\frac{3}{2}xy^2-6xy\)

Thay \(x=0,5;y=1\)vaof  P; dc:

\(P=\frac{3}{2}\cdot0,5-6.0,5=\frac{1}{2}\left(\frac{3}{2}-\frac{12}{2}\right)=\frac{1}{2}\cdot\frac{-9}{2}=-\frac{9}{4}\)

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.`

`Answer:`

\(a)\left(-3x^2y-2xy^2+6\right)+\left(-x^2y+5xy^2-1\right)\)

\(=-3x^2y-2xy^2+6+-x^2y+5xy^2-1\)

\(=\left(-3x^2y-x^2y\right)+\left(-2xy^2+5xy^2\right)+\left(6-1\right)\)

\(=-4x^2y+3xy^2+5\)

\(b)\left(1,6x^3-3,8x^2y\right)+\left(-2,2x^2y-1,6x^3+0,5xy^2\right)\)

\(=1,6x^3-3,8x^2y+-2,2x^2y-1,6x^3+0,5xy^2\)

\(=\left(1,6x^3-1,6x^3\right)+\left(-3,8x^2y+-2,2x^2y\right)+0,5xy^2\)

\(=-6x^2y+0,5xy^2\)

\(c)\left(6,7xy^2-2,7xy+5y^2\right)-\left(1,3xy-3,3xy^2+5y^2\right)\)

\(=6,7xy^2-2,7xy+5y^2-1,3xy+3,3xy^2-5y^2\)

\(=\left(6,7xy^2+3,3xy^2\right)+\left(-2,7xy-1,3xy\right)+\left(5y^2-5y^2\right)\)

\(=10xy^2+-4xy\)

\(=10xy^2-4xy\)

\(d)\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

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

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

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

\(e)\left(x^2+y^2-2xy\right)-\left(x^2+y^2+2xy\right)+\left(4xy-1\right)\)

\(=x^2+y^2-2xy-x^2-y^2-2xy+4xy-1\)

\(=\left(x^2-x^2\right)+\left(y^2-y^2\right)+\left(-2xy-2xy+4xy\right)-1\)

\(=-1\)

a) Ta có: \(M=x^2y+xy^2-5x^2y^2+x^3-2x^2y+6xy^2\)

\(=\left(x^2y-2x^2y\right)+\left(xy^2+6xy^2\right)-5x^2y^2+x^3\)

\(=x^3-x^2y+7xy^2-5x^2y^2\)

Bậc là 4

Ta có: \(N=3x^3+xy+y^2-x^2y^2-2-2xy+7y^2\)

\(=3x^3+\left(xy-2xy\right)+\left(y^2+7y^2\right)-x^2y^2-2\)

\(=3x^2+8y^2-xy-x^2y^2-2\)

Bậc là 4

c) (xy-1).(xy+5)

= x²y²+5xy-xy-5

=x²y²+4xy-5

a) b) d) bạn có thể ghi rõ được ko

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

trả lời hộ với mai thi rồi