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

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

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

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

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

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

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

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

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

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

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

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)

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

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

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

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

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

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

     y\(^2\) - ( x\(^2\) - 2x + 1 )

⇔ y\(^2\) - ( x - 1 )\(^2\)

⇔ ( y - x - 1 ) ( y + x - 1 )