a) x²y+xy+x+1= (x²y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)

b) x²-(a+b)x+ab=x²-ax-bx+ab=(x²-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)

c) ax²+ay-bx²-by=(ax²+ay)-(bx²+by)=a(x²+y)-b(x²+y)=(a-b)(x²+y)

d) ax-2x-a²+2a=(ax-2x)-(a²-2a)=x(a-2)-a(a-2)=(a-2)(x-a)

e) 2x²+4ax+x+2a=(2x²+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)

f) x³+ax²+x+a=(x³+ax²)+(x+a)=x²(x+a)+(x+a)=(x²+1)(x+a)

g: Ta có: \(x^4+2x^3-4x-4\)

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

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

a \(x\cdot\left(a-2\right)-a\left(a-2\right)\)

\(\left(x-a\right)\cdot\left(a-2\right)\)

b ,\(x\left(x+1\right)-a\left(x+1\right)\)

\(\left(x-a\right)\cdot\left(x+1\right)\)

c ,\(2x\left(x+2a\right)+x+2a\)

(2x+1)(x+2a)

d, 2xy+x^2-ax-2ay

x(2y+x)-a(x+2y)

(x-a)(x+2y)

e, x^2(x+a)+x+a

(x^2+1)(x+a)

f, y^2(y+x^2)+z(x^2+y)

(y^2+z)(y+x^2)

2xy - ax + x²-2ay

= (2xy+x²) - (ax + 2ay)

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

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

\(2xy-ax+x^2-2ay\)

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

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

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

2xy - ax + x²-2ay

= (2xy+x²) - (ax + 2ay)

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

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

\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)

a,=x(x−2)b,=2b(x−3y)+a(x−3y)=(a+2b)(x−3y)c,=x(x2+2xy+y2−4)=x[(x+y)2−4]=x(x+y+2)(x+y−2)d,=4−(x+y)2=(2−x−y)(2+x+y)đ,=5(x−y)(x+y)+3(x+y)2=(x+y)(5x−5y+3x+3y)=(x+y)(8x−2y)=2(4x−y)(x+y)e,=3x(2xy−3)b,=x(4x2−4xy+y2−4)=x[(2x−y)2−4]=x(2x−y−2)(2x−y+2)f,=(x+y)2−z2=(x+y−z)(x+y+z)

c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)

d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)

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

f) \(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)

c: \(5x^2+15x+3y+xy\)

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

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

d: \(x^2+6x+9-y^2\)

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

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

e: \(x^2+2x+1-y^2\)

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

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

f: \(x^2-2xy+y^2-9\)

\(=\left(x-y\right)^2-9\)

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

d) ax² + ay - bx² - by

= ( ax² + ay ) - ( bx² + by )

= a ( x² + y ) - b ( x² + y )

=  ( x² + y )( a - b )

c) x²y + xy² - x - y

= ( x²y + xy² ) - ( x + y )

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

= ( x + y ) ( xy - 1 )

b) ax + by + ay + bx

= ax + ay + bx + by

= a ( x + y ) + b ( x + y )

= ( x + y ) ( a + b )

b) Ta có: \(x^2y+xy+x+1\)

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

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

c) Ta có:  \(ax+by+ay+bx\)

\(=a\left(x+y\right)+b\left(x+y\right)\)

\(=\left(x+y\right)\left(a+b\right)\)

d) Ta có: \(x^2-\left(a+b\right)x+ab\)

\(=x^2-ax-bx+ab\)

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

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

a) ko bt làm