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

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

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)

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

\(\dfrac{1}{x-y}-\dfrac{1}{x+y}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y}{\left(x-y\right)\left(x+y\right)}-\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y-x+y+2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2x+2y}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2}{x-y}\)

\(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{4x}{\left(x-y\right)\left(x+y\right)}\)

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

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

\(=\left(2x+1\right)\left(8-2x\right)\)

\(=2\left(4-x\right)\left(2x+1\right)\)

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)

\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)

\(=\left(3x-2\right)\left(3x-6\right)\)

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-3y+4\right)\)

whatterfice

\(3x^2\left(x-1\right)+5x\left(1-x\right)^2\\ =-3x^2\left(1-x\right)+5x\left(1-x\right)\left(1-x\right)\\ =\left(1-x\right)\left(-3x^2+5x-5x^2\right)\\ =x\left(1-x\right)\left(-3x+5-5x\right)\\ =x\left(1-x\right)\left(5-8x\right)\)

\(3\left(x-y\right)^2+9y\left(y-x\right)^2\\ =3\left(x-y\right)\left(x-y\right)+9y\left(y-x\right)\left(y-x\right)\\ =-3\left(x-y\right)\left(y-x\right)+9y\left(y-x\right)\left(y-x\right)\\ =\left(y-x\right)\left(-3x+3y+9y^2-9xy\right)\\ =\left(y-x\right)\left[-3\left(x-y\right)+9y\left(y-x\right)\right]\\ =\left(y-x\right)\left[-3\left(x-y\right)-9y\left(x-y\right)\right]\\ =\left(y-x\right)\left(-3-9y\right)\left(x-y\right)\\ =-3\left(y-x\right)\left(1+3y\right)\left(x-y\right)\)

\(3\left(x-y\right)^2+9y\left(y-x\right)\\ =3\left(x-y\right)\left(x-y\right)-9y\left(x-y\right)\\ =\left(x-y\right)\left(3x-3y-9y\right)\\ =\left(x-y\right)\left(3x-12y\right)\\ =3\left(x-y\right)\left(x-4y\right)\)

\(x^2\left(y-1\right)-4\left(y-1\right)\\ =\left(y-1\right)\left(x^2-4\right)=\left(y-1\right)\left(x-2\right)\left(x+2\right)\)

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

\(x^2\left(y-1\right)-4\left(y-1\right)\\ =\left(x^2-4\right)\left(y-1\right)\\ =\left(x-2\right)\left(x+2\right)\left(y-1\right)\)