xy(x² + y²) + 2 = (x + y)²

=> xy(x + y)² - 2x²y² + 2 - (x + y)² = 0

=> (x + y)²(xy - 1) - 2(x²y² - 1) = 0

=> (x + y)²(xy - 1) - 2[(xy)² - 1²] = 0

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

=> (xy - 1)[(x + y)² - 2(xy + 1)] = 0

=> (xy - 1)(x² + y² - 2) = 0

Vậy...

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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

= (x - y)[(x + y)^2 - x]

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

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

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

\(=\left(x-1\right)\left(x+1\right)\left(1-y\right)\left(1+y\right)-\left(x-1\right)\left(1-y\right)\)

\(=\left(x-1\right)\left(1-y\right)\left[\left(x+1\right)\left(y+1\right)-1\right]\)

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

Lời giải:

a. $xy(x+y)-y(x+y)^2+y^2(x-y)$

$=y(x+y)[x-(x+y)]+y^2(x-y)$

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

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

$=y^2(x-y)-y^2(x+y)=y^2[(x-y)-(x+y)]$

$=y^2(-2y)=-2y^3$

b.

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

$=[x(x+y)^2-y(x+y)^2]-(x^2-xy)$

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

$=(x-y)[(x+y)^2-x]=(x-y)(x^2+2xy+y^2-x)$

a: \(xy\left(x+y\right)-y\left(x+y\right)^2+y^2\left(x-y\right)\)

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

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

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

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

b: \(x\left(x+y\right)^2-y\left(x+y\right)^2+xy-x^2\)

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

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

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

1. x^2 -x-y^2-y= ( x^2-y^2) - ( x+y)= (x+y).(x-y) - ( x+y)= (x+y). ( x-y-1)

2. x^2-y^2+x-y= (x-y).(x+y) + (x-y)= (x-y).(x+y+1)

3. 3x-3y+x^2-y^2= 3.(x-y) + (x-y).(x+y)= (x-y).(3+x+y)

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

-(y^3-2*x*y^2-y^2+x^2*y-2*x*y-x^2)

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

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