( a - x )y³ - ( a - y )x³ - ( x - y )a³

= ay³ - xy³ - ax³ + x³y - ( x - y )a³

= ( x³y - xy³ ) - ( ax³ - ay³ ) - ( x - y )a³

=  xy( x² - y² ) - a( x³ - y³ ) - ( x - y )a³

= xy( x - y )( x + y ) - a( x - y )( x² + xy + y² ) - ( x - y )a³

= ( x - y )[ xy( x + y ) - a( x² + xy + y² ) - a³ ]

= ( x - y )( x²y + xy² - ax² - axy - ay² - a³ )

a) x³ - 8 = ( x - 2 )( x - 12 )

<=> ( x - 2 )( x² + 2x + 4 ) - ( x - 2 )( x - 12 ) = 0

<=> ( x - 2 )( x² + 2x + 4 - x + 12 ) = 0

<=> ( x - 2 )( x² + x + 16 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x^2+x+16=0\end{cases}}\Leftrightarrow x=2\)( vì x² + x + 16 = ( x² + x + 1/4 ) + 63/4 = ( x + 1/2 )² + 63/4 ≥ 63/4 > 0 ∀ x )

b) x²( x² + 4 ) - x² = 4

<=> x²( x² + 4 ) - x² - 4 = 0

<=> x²( x² + 4 ) - ( x² + 4 ) = 0

<=> ( x² + 4 )( x² - 1 ) = 0

<=> \(\orbr{\begin{cases}x^2+4=0\\x^2-1=0\end{cases}}\Leftrightarrow x=\pm1\)( vì x² + 4 ≥ 4 > 0 ∀ x )

Ko khó nè :3, đừng tách ra nhé ! 

a, \(x^3-8=\left(x-2\right)\left(x-12\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)\left(x-12\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-x+12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\Leftrightarrow x=2\)

b, \(x^2\left(x^2+4\right)-x^2=4\Leftrightarrow-x^2\left(-x^2-4\right)-x^2=4\)

\(\Leftrightarrow-x^2\left(4-x^2\right)-x^2=4\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)-x^2-4=0\)

\(\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)+\left(-x^2-4\right)=0\)

\(\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)+\left(2-x\right)\left(2+x\right)=0\)

\(\Leftrightarrow\left(-x^2+1\right)\left(2-x\right)\left(2+x\right)=0\Leftrightarrow x=\pm1;\pm2\)

Check hộ dáp án nhá :), ko chắc lắm nếu khai triển sẽ dễ nhìn hơn đấy. 

( x - y )² + 4x - 4y - 12 < xin phép sửa đề >

= ( x - y )² + 4( x - y ) - 12

Đặt t = x - y

bthuc <=> t² + 4t - 12

            = t² - 2t + 6t - 12

            = t( t - 2 ) + 6( t - 2 )

            = ( t - 2 )( t + 6 )

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

9x² - 4 - ( 3x - 2 )( x + 5 ) = 0

<=> ( 3x - 2 )( 3x + 2 ) - ( 3x - 2 )( x + 5 ) = 0

<=> ( 3x - 2 )( 3x + 2 - x - 5 ) = 0

<=> ( 3x - 2 )( 2x - 3 ) = 0

<=> \(\orbr{\begin{cases}3x-2=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{3}{2}\end{cases}}\)

x³ + 64 + ( x + 4 )( 2x - 3 ) = 0

<=> ( x + 4 )( x² - 4x + 16 ) + ( x + 4 )( 2x - 3 ) = 0

<=> ( x + 4 )( x² - 4x + 16 + 2x - 3 ) = 0

<=> ( x + 4 )( x² - 2x + 13 ) = 0

<=> \(\orbr{\begin{cases}x+4=0\\x^2-2x+13=0\end{cases}}\Leftrightarrow x=-4\)( vì x² - 2x + 13 = ( x² - 2x + 1 ) + 12 = ( x - 1 )² + 12 ≥ 12 > 0 ∀ x )

( x - 3 )( x² + 4x + 9 ) + 2( x² - 9 ) - 10( x - 3 ) = 0

<=> ( x - 3 )( x² + 4x + 9 ) + 2( x - 3 )( x + 3 ) - 10( x - 3 ) = 0

<=> ( x - 3 )( x² + 4x + 9 + 2x + 6 - 10 ) = 0

<=> ( x - 3 )( x² + 6x + 5 ) = 0

<=> ( x - 3 )( x + 1 )( x + 5 ) = 0

<=> x = 3 hoặc x = -1 hoặc x = -5

<=> ( x - 3 )( 

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 )

Gửi nhầm gửi nhầm, k cần trả lời nữa đâu ^ ^''

photomath không nói nhiều

21, \(x^3-4x^2+4x=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\)

22, \(15x^2y+20xy^2-25xy=5xy\left(3x+4y-5\right)\)

23, \(4x^2+8xy-3x-6y=4x\left(x+2y\right)-3\left(x+2y\right)=\left(4x-3\right)\left(x+2y\right)\)

24, \(x^3-6x^2+9x=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)

Tương tự :)) 

21.\(x^3-4x^2+4x\)

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

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

22,\(15x^2y+20xy^2-25xy\)

\(=5xy\left(3x+4y-5\right)\)

23,\(4x^2+8xy-3x-6y\)

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

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

24\(x^3-6x^2+9x\)

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

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

25,\(x^2-xy+x-y\)

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

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

26.\(xy-2x-y^2+2y\)

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

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

27,\(x^2+x-xy-y\)

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

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

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

28,\(x^2+4x-y^2+4\)

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

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

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

29.\(x^2-2xy+y^2-4\)

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

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

16) 2x + 2y - x² - xy = ( 2x + 2y ) - ( x² + xy ) = 2( x + y ) - x( x + y ) = ( x + y )( 2 - x )

17) x² - 2x - 4y² - 4y = ( x² - 4y² ) - ( 2x + 4y ) = ( x - 2y )( x + 2y ) - 2( x + 2y ) = ( x + 2y )( x - 2y - 2 )

18) x²y - x³ - 9y + 9x = ( x²y - x³ ) - ( 9y - 9x ) = x²( y - x ) - 9( y - x ) = ( y - x )( x² - 9 ) = ( y - x )( x - 3 )( x + 3 )

19) x²( x - 1 ) + 16( 1 - x ) = x²( x - 1 ) - 16( x - 1 ) = ( x - 1 )( x² - 16 ) = ( x - 1 )( x - 4 )( x + 4 )

20) 2x² + 3x - 2xy - 3y = ( 2x² - 2xy ) + ( 3x - 3y ) = 2x( x - y ) + 3( x - y ) = ( x - y )( 2x + 3 )

20, \(2x^2+3x-2xy-3y=2x\left(x-y\right)+3\left(x-y\right)=\left(2x+3\right)\left(x-y\right)\)

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

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

18, \(x^2y-x^3-9y+9x=-x\left(x^2-9\right)+y\left(x^2-9\right)=\left(-x-y\right)\left(x^2-9\right)=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)

19, \(x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x^2-16\right)\left(x-1\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\)

a) ( x + 1/2 )² - ( x + 1/2 )( x + 6 ) = 8

<=> ( x + 1/2 )[ ( x + 1/2 ) - ( x + 6 ) ] = 8

<=> ( x + 1/2 )( x + 1/2 - x - 6 ) = 8

<=> ( x + 1/2 ).(-11/2) = 8

<=> x + 1/2 = -16/11

<=> x = -43/22

b) ( x² + 2x )² - 2x² - 4x = 3

<=> ( x² + 2x )² - 2( x² + 2x ) - 3 = 0 (*)

Đặt t = x² + 2x

(*) <=> t² - 2t - 3 = 0

     <=> t² + t - 3t - 3 = 0

     <=> t( t + 1 ) - 3( t + 1 ) = 0

     <=> ( t + 1 )( t - 3 ) = 0

     <=> ( x² + 2x + 1 )( x² + 2x - 3 ) = 0

     <=> ( x + 1 )²( x² - x + 3x - 3 ) = 0

     <=> ( x + 1 )²[ x( x - 1 ) + 3( x - 1 ) ] = 0

     <=> ( x + 1 )²( x - 1 )( x + 3 ) = 0

     <=> x = -1 hoặc x = 1 hoặc x = -3

a, \(\left(5x-2y\right)\left(5x+2y\right)+4y-1=25x^2-4y^2+4y-1\)

\(=\left(5x\right)^2-\left(4y^2-4y+1\right)=\left(5x\right)^2-\left(2y-1\right)^2=\left(5x-2y+1\right)\left(5x+2y-1\right)\)

b, t đag loading :)) phải vận dụng kiến thức đầu có nhóm zô, tí giải choa nhá :) 

Okie :)))