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

\(a.25^2-4a^2+12ab-9b^2\\ =25^2-\left(4a^2+12ab-9b^2\right)\\ =25^2-\left(2a-3b\right)^2\\ =\left(25-2a+3b\right)\left(25+2a-3b\right)\\ b.x^3+x^2y-xy^2-y^3\\ =x^2\left(x+y\right)-y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-y^2\right)\\ =\left(x+y\right)\left(x+y\right)\left(x-y\right)\\ =\left(x+y\right)^2\left(x-y\right)\)

a: Ta có: \(25x^2-4a^2+12ab-9b^2\)

\(=25x^2-\left(2a-3b\right)^2\)

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

b: Ta có: \(x^3+x^2y-xy^2-y^3\)

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

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

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

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

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

3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)

4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

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

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

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

a)x²-xy-x+y

=(x²-x)-(xy-y)

=x(x-1)-y(x-1)

=(x-1)(x-y)

b) xy+4-x²+2y

=(4-x²)+(xy+2y)

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

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

c) xy+y-2(x+1)

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

=(x+1)(y-2)

d) 5(x-y)+ax-ay

=5(x-y)+a(x-y)

=(x-y)(5+a)

#H

Trả lời:

a, x² - xy - x + y

= ( x² - xy ) - ( x - y )

= x ( x - y ) - ( x - y )

= ( x - y ) ( x - 1 )

b, xy + 4 - x² + 2y 

= ( xy + 2y ) - ( x² - 4 )

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

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

c, xy + y - 2 ( x + 1 ) 

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

= ( x + 1 ) ( y - 2 )

d, 5 ( x - y ) + ax - ay

= 5 ( x - y ) + a ( x - y )

= ( 5 + a ) ( x - y )

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

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

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

d)\(5\left(x-y\right)+ax-ay=5\left(x-y\right)+\left(ax-ay\right)=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)

1.A

2.C

3.B

4.C

a

c

b

c

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

a) \(4\left(2-x\right)^2+xy-2y\)

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

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

\(=\left(x-2\right)\left(4x-8\right)+y\left(x-2\right)\)

\(=\left(x-2\right)\left(4x-8+x-2\right)\)

\(=\left(x-2\right)\left(5x-10\right)\)

\(=5\left(x-2\right)^2\)

a, \(=4\left(x-2\right)^2+y\left(x-2\right)=\left(x-2\right)\left(4x-8+y\right)\)

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

c, \(=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\)

d, không phân tích được

c, x²y - xy² - 3x + 3y
= xy(x-y) - 3(x-y)
= (x-y)(x-3)