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 

dễ thì làm dùm đi, con pk ba giỏi r

2. => (2x-5)²(2x+5)²-9(2x-5)²

=> (2x-5)²(2x-4) =>2(x-2)(2x-5)²

bn chép lại đề nhé

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

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

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

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

c/ \(=2a^2+2b^2-2c^2+4ab=2\left[\left(a^2+b^2+2ab\right)-c^2\right]\)

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

d/ \(=\left(4x^2-25\right)^2-9\left(4x^2-20x+25\right)\)

\(=\left(4x^2-25\right)^2-9\left(4x^2+25\right)+180x\)

tới đây bạn đặt a= 4x^2 -25 rồi làm típ nha, mình lười quá >< 

e/ tương tự câu d nha bạn

f/ \(=a^4\left(a^2-1\right)+2a^2\left(a+1\right)\)

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

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

g/   đặt \(a=3x^2+3x+2\) khi đó biểu thức trở thành

\(a^2-\left(a+4\right)^2=a^2-a^2-8a-16\)

\(=-8a-16=-8\left(3x^2+3x+2-8\right)=-8\left(3x^2+3x-6\right)\)

\(=-24\left(x^2+x-2\right)=-24\left(x-1\right)\left(x+2\right)\)

xong rùi nha bn. Chúc bn hc tốt (xin lỗi tại có mấy câu mình lười nha)

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

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

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

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

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

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

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

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

a/ (x - 1)(x - √3 + 2)(x + √3 + 2)

a ) \(x^3+3x^2-3x+1\)

    \(=x^3-3x+3x^2-1\)

     \(=\left(x-1\right)^3\)

c/ (x + 1)(x + \(\frac{\sqrt{21}-5}{2}\))(x + \(\frac{-\sqrt{21}-5}{2}\))

a)  4(2x-3)^2-9(4x^2-9)^2

=[2(2x-3)]^2-[3(4x^2-9)]^2

=(4x-6)^2-(12x^2-27)^2

=(4x-6+12x^2-27)(4x-6-12x^2+27)

=(12x^2+4x-33)(4x-12x^2+21)

b)  a^6-a^4+2a^3+2a^2

=a^4(a^2-1)+2a^2(a+1)

=a^4(a+1)(a-1)+2a^2(a+1)

=(a+1)[(a^4)(a-1)+2a^2]

=(a+1)(a^5+a^4+2a^2)

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

\(=\left(2x+1\right)\left(8-2x\right)\)

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)

\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)

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

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

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

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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

whatterfice

Bài 2;

\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)

\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)

\(x.\left(x-3\right)-x+3=0\)

\(x.\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

\(x^3-3x^2+3x-1=0\)

\(\left(x-1\right)^3=0\)( hằng đẳng thức số 5 )

\(\Rightarrow x=1\)

Vậy \(x=1\)