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

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

\(=3x^2+3y^2-2.1\left(x^2-xy+y^2\right)\)

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

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

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

\(=x^3+y^3+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2.1\)

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

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

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

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

`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

`b, x(x-2y) - y(y^2-2x)`

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

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)

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

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

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)

\(x+y=1\)

\(\Leftrightarrow\)\(\left(x+y\right)^2=1\)

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

\(x+y=1\)

\(\Leftrightarrow\)\(\left(x+y\right)^3=1\)

\(\Leftrightarrow\)\(x^3+y^3=1-3xy\)

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

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

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

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

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

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

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

\(=1-6x^4y^2\)

mới ra đc đến đây

a: \(A=2\left(x+y\right)+3xy\left(x+y\right)+5x^2y^2\left(x+y\right)=0\)

b: \(B=3xy\left(x+y\right)+2x^2y\left(x+y\right)=0\)

\(\Rightarrow\)A=2(x+y)+3xy(x+y)+5x²y²(x+y)

Thay x+y=0 vào A

\(\Rightarrow\)A=0

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

\(N=x^3+y^3+6x^2y^2+3xy\left[\left(x+y\right)^2-2xy\right]\)

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

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

\(N=\left(x+y\right)^2\)

\(N=1\)

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

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

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

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

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

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

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

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

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

\(C=1-3xy+3xy\)(vì x+y=1)

\(C=1\)

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

\(D=2\left(1^3-3xy\right)-3\left(1^2-2xy\right)\)(vì x+y=1)

\(D=2-6xy-3+6xy\)

\(D=-1\)

a: \(P=2x^2+3xy+y^2=\left(2x+y\right)\left(x+y\right)\)

\(=\left(2\cdot\dfrac{-1}{2}+\dfrac{2}{3}\right)\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)\)

\(=\dfrac{-1}{3}\cdot\dfrac{1}{6}=-\dfrac{1}{18}\)

d: \(Q=\dfrac{-1}{3}x^4y^2=\dfrac{-1}{3}\cdot16\cdot\dfrac{1}{16}=-\dfrac{1}{3}\)