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

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

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

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

x² - x - y² - y

= (x² - y²) - (x + y)

= (x - y).(x + y) - (x + y)

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

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

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

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

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

x^2-X-Y^2-Y=(X^2-Y^2)-(X+Y)=(X-Y)(X+Y)-(X+Y)=(X+Y)(X-Y-1)

x²-x-y²-y

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

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

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

********nha

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

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

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

\(=\left[\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]-4y^2\)

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

\(=\left(2y\right)^2-\left(2y\right)^2=0\)

Sửa:

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

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

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

\(=2x^2-6y^2+4xy\)

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

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

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

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

\(4x\left(x-y\right)+7y\left(x-y\right)+y^2\left(y-x\right)=\left(y-x\right)\left(y^2-7y-4x\right)\)

4x( y - x ) + 7y( x - y ) + y²( y - x )

= 4x( y - x ) - 7y( y - x ) + y²( y - x )

= ( y - x )( 4x - 7y + y² ) 

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

Bài làm:

Ta có: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)

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

Học tốt!!!!

Ta có : 

\(9\left(x-y\right)^2-4\left(x+y\right)^2=9x^2-18xy+9y^2-4x^2-8xy-4y^2\)

\(=5x^2-26xy+5y^2==\left(5x-y\right)\left(x-5y\right)\)