a) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)

\(=\left(4x^2-25\right)^2-\left(6x-15\right)^2\)

\(=\left(4x^2-25-6x+15\right)\left(4x^2-25+6x-15\right)\)

\(=\left(4x^2-6x-10\right)\left(4x^2+6x-40\right)\)

\(=\left(4x^2+4x-10x-10\right)\left(4x^2+16x-10x-40\right)\)

\(=\left[4x\left(x+1\right)-10\left(x+1\right)\right]\left[4x\left(x+4\right)-10\left(x+4\right)\right]\)

\(=\left(4x-10\right)\left(x+1\right)\left(4x-10\right)\left(x+4\right)\)

\(=\left(4x-10\right)^2\left(x+1\right)\left(x+4\right)\)

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

b) \(a^6-a^4+2a^3+2a^2\)

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

\(=a^2\left(a^4+a^3-a^3-a^2+2a+2\right)\)

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

\(=a^2\left(a+1\right)\left(a^3-a^2+2\right)\)

a) ( x² - 25 )² - ( x - 5 )²

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

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

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

= ( x - 5 )²( x + 4 )( x + 6 )

b) ( 4x² - 25 )² - 9( 2x - 5 )²

= ( 4x² - 25 )² - 3²( 2x - 5 )²

= ( 4x² - 25 )² - ( 6x - 15 )² 

= [ ( 4x² - 25 ) - ( 6x - 15 ) ][ ( 4x² - 25 ) + ( 6x - 15 ) ]

= ( 4x² - 25 - 6x + 15 )( 4x² - 25 + 6x - 15 )

= ( 4x² - 6x - 10 )( 4x² + 6x - 40 )

= ( 4x² + 4x - 10x - 10 )( 4x² + 16x - 10x - 40 )

= [ 4x( x + 1 ) - 10( x + 1 ) ][ 4x( x + 4 ) - 10( x + 4 ) ]

= ( x + 1 )( 4x - 10 )( x + 4 )( 4x - 10 )

= ( 4x - 10 )²( x + 1 )( x + 4 )

c) 4( 2x - 3 )² - 9( 4x² - 9 )²

= 2²( 2x - 3 )² - 3²( 4x² - 9 )²

= ( 4x - 6 )² - ( 12x² - 27 )²

= [ ( 4x - 6 ) - ( 12x² - 27 ) ][ ( 4x - 6 ) + ( 12x² - 27 ) ]

= ( 4x - 6 - 12x² + 27 )( 4x - 6 + 12x² - 27 )

= ( -12x² + 4x + 21 )( 12x² + 4x - 33 )

= ( -12x² + 18x - 14x + 21 )( 12x² - 18x + 22x - 33 )

= [ -12x( x - 3/2 ) - 14( x - 3/2 ) ][ 12x( x - 3/2 ) + 22( x - 3/2 ) ]

= ( x - 3/2 )( -12x - 14 )( x - 3/2 )( 12x + 22 )

= ( x - 3/2 )²( -12x - 14 )( 12x + 22 )

d) a⁶ - a⁴ + 2a³ + 2a²

= a²( a⁴ - a² + 2a + 2 )

= a²( a⁴ - 2a³ + 2a³ + 2a² - 4a² + a² + 4a - 2a + 2 )

= a²[ ( a⁴ - 2a³ + 2a² ) + ( 2a³ - 4a² + 4a ) + ( a² - 2a + 2 ) ]

= a²[ a²( a² - 2a + 2 ) + 2a( a² - 2a + 2 ) + 1( a² - 2a + 2 ) ]

= a²( a² + 2a + 1 )( a² - 2a + 2 )

= a²( a + 1 )²( a² - 2a + 2 )

e) ( 3x² + 3x + 2 )² - ( 3x² + 3x - 2 )²

= [ ( 3x² + 3x + 2 ) - ( 3x² + 3x - 2 ) ][ ( 3x² + 3x + 2 ) + ( 3x² + 3x - 2 ) ]

= ( 3x² + 3x + 2 - 3x² - 3x + 2 )( 3x² + 3x + 2 + 3x² + 3x - 2 )

= 4( 6x² + 6x ) 

= 4.6x( x + 1 )

= 24( x + 1 )

e) là 24x( x + 1 ) nhé mình đánh thiếu 

a: \(x^2-9-x^2\left(x^2-9\right)\)

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

\(=\left(x^2-9\right)\left(1-x^2\right)\)

\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)

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

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

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

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

c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

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

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

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

d: \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

e: \(3x^2-4x-4\)

\(=3x^2-6x+2x-4\)

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

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

g: \(x^4+64y^4\)

\(=x^4+16x^2y^2+64y^4-16x^2y^2\)

\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)

\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)

h: \(a^2+b^2+2a-2b-2ab\)

\(=a^2-2ab+b^2+2a-2b\)

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

i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)

\(=\left(x+1-y+3\right)^2\)

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

k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)

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

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

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

7, \(10ax-5ay-2x+y=5a\left(2x-y\right)-\left(2x-y\right)=\left(5a-1\right)\left(2x-y\right)\)

8, xem lại đề bạn nhé

9, \(4x^2-y^2+8y-16=4x^2-\left(y^2-8y+16\right)=4x^2-\left(y-4\right)^2\)

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

Trả lời:

6, x²y + xy² - 4x - 4y = ( x²y + xy² ) - ( 4x + 4y ) = xy ( x + y ) - 4 ( x + y ) = ( x + y )( xy - 4 )

7, 10ax - 5ay - 2x + y = ( 10ax - 5ay ) - ( 2x - y ) = 5a ( 2x - y ) - ( 2x - y ) = ( 2x - y )( 5a - 1 ) 

8, Sửa đề: x³ - 2x² + 2x - 4 = ( x³ - 2x² ) + ( 2x - 4 ) = x² ( x - 2 ) + 2 ( x - 2 ) = ( x - 2 )( x² + 2 )

9, 4x² - y² + 8y - 16 = 4x² - ( y² - 8y + 16 ) = 4x² - ( y - 4 )² = ( 2x - y + 4 )( 2x + y - 4 )

Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

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

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

 \(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)  

= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)

1.

\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x+4}{x-3}\)

b.

\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)

\(\Rightarrow x=10\) (thỏa mãn)

2.

\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)

\(a^5-a\)

\(=a\left(a^4-1\right)\)

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

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