a) \(9\left(2x-3\right)^2-4\left(x+1\right)^2\)

\(=\left[3\left(2x-3\right)-2\left(x+1\right)\right]\left[3\left(2x-3\right)+2\left(x+1\right)\right]\)

\(=\left(6x-9-2x-2\right)\left(6x-9+2x+2\right)\)

\(=\left(4x-11\right)\left(8x-7\right)\)

b) \(\left(x^2+4y^2-20\right)-16\left(xy-4\right)^2\)

\(=\left[\left(x^2-4xy+4y^2\right)-4\right]\left[\left(x^2+4xy+4y^2\right)-36\right]\)

\(=\left[\left(x-2y\right)^2-4\right]\left[\left(x+2y\right)^2-36\right]\)

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

a. 9 ( 2x - 3 )² - 4 ( x + 1 )²

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

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

= ( 6x - 9 - 2x - 2 ) ( 6x - 9 + 2x + 2 )

= ( 4x - 11 ) ( 8x - 7 )

b. ( x² + 4y² - 20 )² - 16 ( xy - 4 )²

= ( x² + 4y² - 20 )² - [ 4 ( xy - 4 ) ]²

= [ x² + 4y² - 20 - 4 ( xy - 4 ) ] [ x² + 4y² - 20 + 4 ( xy - 4 ) ]

= ( x² + 4y² - 20 - 4xy + 16 ) ( x² + 4y² - 20 + 4xy - 16 )

 = ( x² + 4y² - 4xy - 4 ) ( x² + 4y² + 4xy - 36 )

= [ ( x - 2y )² - 2² ] [ ( x + 2y )² - 6² ]

= ( x - 2y - 2 ) ( x - 2y + 2 ) ( x + 2y - 6 ) ( x + 2y + 6 )

a) 9( 2x - 3 )² - 4( x + 1 )²

= 3²( 2x - 3 )² - 2²( x + 1 )²

= [ 3( 2x - 3 ) ]² - [ 2( x + 1 ) ]²

= ( 6x - 9 )² - ( 2x + 2 )²

= [ ( 6x - 9 ) - ( 2x + 2 ) ][ ( 6x - 9 ) + ( 2x + 2 ) ]

= ( 6x - 9 - 2x - 2 )( 6x - 9 + 2x + 2 )

= ( 4x - 11 )( 8x - 7 )

b) ( x² + 4y² - 20 )² - 16( xy - 4 )²

= ( x² + 4y² - 20 )² - 4²( xy - 4 )²

= ( x² + 4y² - 20 )² - [ 4( xy - 4 ) ]²

= ( x² + 4y² - 20 )² - ( 4xy - 16 )²

= [ ( x² + 4y² - 20 ) - ( 4xy - 16 ) ][ ( x² + 4y² - 20 ) + ( 4xy - 16 ) ]

= ( x² + 4y² - 20 - 4xy + 16 )( x² + 4y² - 20 + 4xy - 16 )

= [ ( x² - 4xy + 4y² ) - 4 ][ ( x² + 4xy + 4y² ) - 36 ]

= [ ( x - 2y )² - 2² ][ ( x + 2y )² - 6² ]

= ( x - 2y - 2 )( x - 2y + 2 )( x + 2y - 6 )( x + 2y + 6 )

a) Ta có: \(a^3y^3+125\)

\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)

b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)

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

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

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

\(9\left(x-3y\right)^2-25\left(2x+y\right)^2\)

\(=\left[3\left(x-3y\right)\right]^2-\left[5\left(2x+y\right)\right]^2\)

\(=\left(3x-9y\right)^2-\left(10x+5y\right)^2\)

\(=\left[3x-9y+10x+5y\right]\left[3x-9y-\left(10x+5y\right)\right]\)

\(=\left(13x-4y\right)\left(-7x-14y\right)\)

\(=-7\left(x+2y\right)\left(13x-4y\right)\)

9(x - 3y)² - 25(2x + y)²

= 3².(x - 3y)² - 5².(2x + y)²

= (3x - 9y)² - (10x + 5y)²

= (3x - 9y - 10x - 5y)(3x - 9y + 10x + 5y)

= (-7x - 14y)(13x - 4y)

= -7(x + 2y)(13x - 4y)

sorry ! mk đang học lớp 6

đề sai rùi phải là : \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)

\(=>\left[6\left(x-y\right)\right]^2-\left[5\left(2x-1\right)\right]^2=\left[6\left(x-y\right)-5\left(2x-1\right)\right]\left[6\left(x-y\right)+5\left(2x-1\right)\right]\)

\(=>\left(6x-6y-10x+5\right)\left(6x-6y+10x-5\right)=\left(5-4x-6y\right)\left(16x-6y-5\right)\)

Áp dụng HDT : x^2 -y^2 =(x-y) (x+y)

Ủng hộ = 1 cái t i c k nha cảm ơn

\(1,=3xy\left(x^2+2xy+y^2\right)=3xy\left(x+y\right)^2\\ 2,=7xy\left(2x-3y+4xy\right)\\ 3,=\left(x-1\right)\left(x^2-4\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\\ 4,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ 5,=\left(b-c\right)\left(8a-6b\right)=2\left(4a-3b\right)\left(b-c\right)\\ 6,=\left(x-1\right)\left(x^2-16\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\\ 7,=x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(x+5\right)\\ 8,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\\ 9,=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\\ 10,=\left(x-1\right)^2-4y^2=\left(x-2y-1\right)\left(x+2y-1\right)\)