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

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

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

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

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

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

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

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

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

 tại sao lại có +1 vậy ạ

a) \(xy-y^2-x+y\)

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

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

b) \(xy-x^2-x+y\)

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

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

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+3\right)\left(x+2\right)\)

a: =(x+y)^2*(x-y)+x(y-x)

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

dấu * là j vậy bạn

b: \(3x+3y-x^2-2xy-y^2\)

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

\(=\left(x+y\right)\left(3-x-y\right)\)

b) 3x+3y-x²-2xy-y²

=3(x+y)-(x+y)²

=(x+y)(3-x-y)