a) ( x - 1 )( 2x + 1 ) + 3( x - 1 )( x + 2 )( 2x + 1 )

= ( x - 1 )( 2x + 1 )[ 1 + 3( x + 2 ) ]

= ( x - 1 )( 2x + 1 )( 1 + 3x + 6 )

= ( x - 1 )( 2x + 1 )( 3x + 7 )

b) ( 6x + 3 ) - ( 2x - 5 )( 2x + 1 )

= 3( 2x + 1 ) - ( 2x - 5 )( 2x + 1 )

= ( 2x + 1 )[ 3 - ( 2x - 5 ) ]

= ( 2x + 1 )( 3 - 2x + 5 )

= ( 2x + 1 )( 8 - 2x )

= 2( 2x + 1 )( 4 - x )

c) ( x - 5 )² + ( x + 5 )( x - 5 ) - ( 5 - x )( 2x + 1 )

= ( x - 5 )² + ( x + 5 )( x - 5 ) + ( x - 5 )( 2x + 1 )

= ( x - 5 )[ ( x - 5 ) + ( x + 5 ) + ( 2x + 1 ) ]

= ( x - 5 )( x - 5 + x + 5 + 2x + 1 )

= ( x - 5 )( 4x + 1 )

d) ( 3x - 2 )( 4x - 3 ) - ( 2 - 3x )( x - 1 ) - 2( 3x - 2 )( x + 1 )

= ( 3x - 2 )( 4x - 3 ) + ( 3x - 2 )( x - 1 ) - 2( 3x - 2 )( x + 1 )

= ( 3x - 2 )[ ( 4x - 3 ) + ( x - 1 ) - 2( x + 1 ) ]

= ( 3x - 2 )( 4x - 3 + x - 1 - 2x - 2 )

= ( 3x - 2 )( 3x - 6 )

= 3( 3x - 2 )( x - 2 )

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Bài 1.

a) x³ + 2x² - 3x - 6 = ( x³ + 2x² ) - ( 3x + 6 ) = x²( x + 2 ) - 3( x + 2 ) = ( x + 2 )( x² - 3 )

b) ( x - 9 )( x - 7 ) + 1 = x² - 16x + 63 + 1 = x² - 16x + 64 = ( x - 8 )²

c) ( x² + x - 1 )² + 4x² + 4x 

= ( x² + x - 1 )² + 4( x² + x ) (1)

Đặt t = x² + x

(1) <=> ( t - 1 )² + 4t

       = t² - 2t + 1 + 4t

       = t² + 2t + 1

       = ( t + 1 )²

       = ( x² + x + 1 )²

d) ( x² + y² - 17 )² - 4( xy - 4 )²

= ( x² + y² - 17 )² - 2²( xy - 4 )²

= ( x² + y² - 17 )² - [ 2( xy - 4 ) ]²

= ( x² + y² - 17 )² - ( 2xy - 8 )²

= [ ( x² + y² - 17 ) - ( 2xy - 8 ) ][ ( x² + y² - 17 ) + ( 2xy - 8 ) ]

= ( x² + y² - 17 - 2xy + 8 )( x² + y² - 17 + 2xy - 8 )

= [ ( x² - 2xy + y² ) - 17 + 8 ][ ( x² + 2xy + y² ) - 17 - 8 ]

= [ ( x - y )² - 9 ][ ( x + y )² - 25 ]

= [ ( x - y )² - 3² ][ ( x + y )² - 5² ]

= ( x - y - 3 )( x - y + 3 )( x + y - 5 )( x + y + 5 )

Bài 2.

ĐK : x, y ∈ Z

a) x + 2y = xy + 2

<=> x + 2y - xy - 2 = 0

<=> ( x - xy ) - ( 2 - 2y ) = 0

<=> x( 1 - y ) - 2( 1 - y ) = 0

<=> ( 1 - y )( x - 2 ) = 0

+) Nếu 1 - y = 0 => y = 1 và nghiệm đúng với mọi x ∈ Z

+) Nếu x - 2 = 0 => x = 2 và nghiệm đúng với mọi y ∈ Z 

Vậy phương trình có hai nghiệm 

1. \(\hept{\begin{cases}y=1\\\forall x\inℤ\end{cases}}\); 2. \(\hept{\begin{cases}x=2\\\forall y\inℤ\end{cases}}\)

b) xy = x + y

<=> xy - x - y = 0

<=> ( xy - x ) - ( y - 1 ) - 1 = 0

<=> x( y - 1 ) - ( y - 1 ) = 1

<=> ( y - 1 )( x - 1 ) = 1

Ta có bảng sau : 

Các nghiệm trên đều thỏa mãn ĐK

Vậy ( x ; y ) = { ( 2 ; 2 ) , ( 0 ; 0 ) }

a,\(x^3+2x^2-3x-6\)

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

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

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

b,\(\left(x-9\right)\left(x-7\right)+1\)

\(=x^2-7x-9x+63+1\)

\(=x^2-16x+64\)

\(=\left(x-8\right)^2\)

a) x( x + 2 )( x + 3 )( x + 5 ) + 5

= [ x( x + 5 ) ][ ( x + 2 )( x + 3 ) ] + 5

= ( x² + 5x )( x² + 5x + 6 ) + 5 (1)

Đặt t = x² + 5x

(1) <=> t( t + 6 ) + 5

       = t² + 6t + 5

       = t² + t + 5t + 5 

       = t( t + 1 ) + 5( t + 1 )

       = ( t + 1 )( t + 5 )

       = ( x² + 5x + 1 )( x² + 5x + 5 )

b) 6x² - 5xy + y² = 6x² - 3xy - 2xy + y² = 3x( 2x - y ) - y( 2x - y ) = ( 2x - y )( 3x - y )

a,\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+5\)

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

\(=\left(x^2+5x\right)\left(x^2+5x+6\right)+5\)(*)

Đặt \(a=x^2+5x\)ta đc:

(*)=\(a\left(a+6\right)+5\)

\(=a^2+6a+5\)

\(=a^2+a+5a+5\)

\(=a\left(a+1\right)+5\left(a+1\right)\)

\(=\left(a+5\right)\left(a+1\right)\)

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

b,\(6x^2-3xy-2xy+y^2\)

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

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

À HÁ 

• JAIUUUUUUUUUUUUH

a) 5x² - 5xy + 7y - 7x = ( 5x² - 5xy ) - ( 7x - 7y ) = 5x( x - y ) - 7( x - y ) = ( x - y )( 5x - 7 )

b) x² - y² + 2x + 1 = ( x² + 2x + 1 ) - y² = ( x + 1 )² - y² = ( x - y + 1 )( x + y + 1 )

c) 3x² + 6xy + 3y² - 3z² = 3( x² + 2xy + y² - z² ) = 3[ ( x² + 2xy + y² ) - z² ] = 3[ ( x + y )² - z² ] = 3( x + y - z )( x + y + z )

d) ab( x² + y² ) + xy( a² + b² ) = abx² + aby² + a²xy + b²xy

                                                = ( a²xy + abx² ) + ( aby² + b²xy )

                                                = ax( ay + bx ) + by( ay + bx )

                                                = ( ay + bx )( ax + by )

a) 16x² - ( x² + 4 )²

= ( 4x )² - ( x² + 4 )²

= [ 4x - ( x² + 4 ) ][ 4x + ( x² + 4 ) ]

= ( -x² + 4x - 4 )( x² + 4x + 4 )

= [ -( x² - 4x + 4 ) ]( x + 2 )²

= [ -( x - 2 )² ]( x + 2 )²

b) ( x + y )³ + ( x - y )³

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

= ( x + y + x - y )[ x² + 2xy + y² - ( x² - y² ) + x² - 2xy + y² ]

= 2x( 2x² + 2y² - x² + y²

= 2x( x² + 3y² )

16 = 4 ‧ 4 = 8 ‧ 2

Mà 4 - 4 = 0 , 8 - 2 = 6

x = 8 , y = 2

\(x^3-y^3=8^3-2^3=512-8=504\)

\(\Rightarrow x^3-y^3=504\)

x - y = 6

=> ( x - y )² = 36

=> x² - 2xy + y² = 36

=> x² + y² - 32 = 36

=> x² + y² = 68

Ta có x³ - y³ = ( x - y )( x² + xy + y² )

                     = 6.( 68 + 16 )

                     = 6.84 = 504