\(x+y=1\)

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

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

\(x+y=1\)

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

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

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

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

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

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

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

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

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

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

mới ra đc đến đây

`a, = 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:
\(5^2-3^3=-3\)

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

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

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

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

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

\(N=1\)

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

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

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

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

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

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)

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

\(A=x^3-y^3-21xy\)

\(A=\left(x-y\right).\left(x^2+xy+y^2\right)-21xy\)

\(A=7.\left(x^2+xy+y^2\right)-21xy\)

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

\(A=7.\left(x^2+2xy+y^2+2xy\right)\)

\(A=7.\text{[}\left(x+y\right)^2+2xy\text{]}\)

\(A=7.\left(7^2+2xy\right)\)

\(A=7^3+14xy\)

Ngáo rồi @@

\(\)

\(A=x^3-y^3-21xy\)

\(\Rightarrow A=\left(x-y\right)\left(x^2+xy+y^2\right)-21xy\)

\(\Rightarrow A=7\left(x^2+xy+y^2\right)-21xy\)

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

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

\(\Rightarrow A=7\left(x-y\right)^2\)

\(\Rightarrow A=7.7^2\)

\(\Rightarrow A=7.49\)

\(\Rightarrow A=343\)

ak trách j :V lộn dấu :V 

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