a ) xy + 1 - x - y

= x ( y - 1 ) + 1 - y

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

= ( x - 1 ) ( y - 1 )

b ) x² + ab + ac + bx

= ( b + x )x + a( b + c )

= ( b + x )1x + 1a( b + c )

= ( b + x ) ( x + a ) ( b + c )

c ) ax + bx - cx + a + b - c

= ( a + b - c )x + a + b - c

= ( a + b - c )x + ( a + b - c )1

= ( a + b - c ) ( x + 1 )

A = 4acx + 4bcx + 4ax + 4bx ( đã sửa '-' )

= 4x( ac + bc + a + b )

= 4x[ c( a + b ) + ( a + b ) ]

= 4x( a + b )( c + 1 )

B = ax - bx + cx - 3a + 3b - 3c

= x( a - b + c ) - 3( a - b + c )

= ( a - b + c )( x - 3 )

C = 2ax - bx + 3cx - 2a + b - 3c

= x( 2a - b + 3c ) - ( 2a - b + 3c )

= ( 2a - b + 3c )( x - 1 )

D = ax - bx - 2cx - 2a + 2b + 4c

= x( a - b - 2c ) - 2( a - b - 2c )

= ( a - b - 2c )( x - 2 )

E = 3ax² + 3bx² + ax + bx + 5a + 5b

= 3x²( a + b ) + x( a + b ) + 5( a + b )

= ( a + b )( 3x² + x + 5 )

F = ax² - bx² - 2ax + 2bx - 3a + 3b

= x²( a - b ) - 2x( a - b ) - 3( a - b )

= ( a - b )( x² - 2x - 3 )

= ( a - b )( x² + x - 3x - 3 )

= ( a - b )[ x( x + 1 ) - 3( x + 1 ) ]

= ( a - b )( x + 1 )( x - 3 )

\(ax^2-3axy+bx-3by\\ =x\left(ax+b\right)-3y\left(ax+b\right)\\ =\left(x-3y\right)\left(ax+b\right)\)

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

\(2ax^3+6ax^2+6ax+18a\\ =2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\)

\(10xy^2-5by^2+2ax-ab\\ =5y^2\left(2x-b\right)+a\left(2x-b\right)\\ =\left(5y^2+a\right)\left(2x-b\right)\)

\(ax-bx+cx-3a+3b-3c\\ =x\left(a-b+c\right)-3\left(a-b+c\right)\\ =\left(x-3\right)\left(a-b+c\right)\)

\(a,=x^2-mx-nx+mn=x\left(x-m\right)-n\left(x-m\right)=\left(x-n\right)\left(x-m\right)\\ b,=a\left(x-y\right)-b\left(x-y\right)+\left(a-b\right)\\ =\left(x-y\right)\left(a-b\right)+\left(a-b\right)=\left(a-b\right)\left(x-y+1\right)\)

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

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

a) \(x^2-\left(m+n\right)x+mn=\left(x^2-n\cdot x\right)-\left(m\cdot x-m\cdot n\right)=x\left(x-n\right)-m\left(x-n\right)=\left(x-m\right)\left(x-n\right)\)

b) \(ax+by+a-bx-ay-b\)

   \(=\left(ax-ay+a\right)-\left(bx-by+b\right)\)

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

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

ax² - ax + bx²  -bx + a + b

= (ax²+ bx² ) - (ax + bx) + (a + b)

=x² (a + b) - x(a + b) + (a + b)

= (x² - x + 1)(a + b)

ax² - ax + bx² - bx + a + b

= ( ax² + bx² ) - ( ax + bx ) + ( a + b )

= x²( a + b ) - x( a + b ) + ( a + b )

= ( a + b )( x² - x + 1 )

\(ax^2-ax+bx^2-bx+a+b\)

\(=ax\left(x-1\right)+bx\left(x-1\right)+\left(a+b\right)\)

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

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

Giả sử x₁,x₂ là 2 nghiệm phân biệt của đa thức P(x)=ax²+bx+c trong đó a khác 0,c khác 0.Hãy tìm nghiệm của đa thức             Q(x)=cx²+bx+a theo x1,x2

d)\(x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)\)

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

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

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

f)\(ax^2+ay-bx^2-by=a\left(x^2+y\right)-b\left(x^2+y\right)\)

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