e) Ta có: \(a^3-a^2-a+1\)

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

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

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

f) Ta có: \(x^3-2xy-x^2y+2y^2\)

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

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

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

b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

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

d) Đề sai ko ???

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

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

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

\(b,=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

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

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

\(e,=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)\)

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

a)\(\left(3x+2\right)\left(2x-3\right)=6x^2-5x-6\)

b) viết lại đề nhin (dùng f(x) viết mới rõ ra dduocj) ko phải dùng {[(...)]} cho chuẩn vào

c) \(\left(x-2\right)^3-x^2.\left(x-6\right)=x^3-3.x.2\left(x-2\right)-8-x^3+6x^2\)

\(=x^3-6x^2+12x-8-x^3+6x^2=12x-8=4\Rightarrow x=1\)

Em lớp 5 thôi nhưng em nghĩ bài này nhân vế bình thường thôi ạ 

Ví dụ câu a : \(\left(2x^2+x-1\right)\left(-3x^2-7x-5\right)\)

\(=-6x^4-17x^3-14x^2+2x+5\)

Tương tự chị nhé 

Câu b : 

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

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

\(=\)\(2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\) 

Câu b : 

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

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

\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)

\(=2x^3+y^3-x^2y-2xy^2\)

Câu c : 

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

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

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

\(=18x^3+9x^4+18x+9x^2-12x^4-6x^5-12x^2-6x^3\)

\(=12x^3-3x^4-3x^2+18x-6x^5\) 

Tách ra mỗi câu một lần.

Dài quá không ai làm đâu.

Nhìn nản lắm.

Câu 3: 

a: \(49^2=2401\)

b: \(51^2=2601\)

c: \(99\cdot100=9900\)

1.a) (4x - 6y)² - (8xy - 5)² = (4x - 6y - 8xy + 5)(4x - 6y + 8xy - 5)

   b) 16x² - 49y² = (4x)² - (7y)² = (4x - 7y)(4x + 7y)

   c) 36x² + 60x + 25 = (6x)² + 2.6x.5 + 5² = (6x + 5)²

   d) (2x - y)(x - y) - (3y - 4x)² + (y - 2x)(2y - 3x) = (y - 2x)(y - x) + (y - 2x)(2y - 3x) - (3y - 4x)²

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

2.M = (3x - 4)(9x² - 12x + 16) + (6x - 8)² = (3x - 4)[(3x)² - 2.3x.4 + 4²] + [2(3x - 4)]² = (3x - 4)(3x - 4)² + 4(3x - 4)²

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

a) =(4x-6y-8xy+3)(4x-6y+8xy-3)

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

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

câu a )Phan Thanh Tịnh sai, tui sửa lại 

a ) ( x - 2 )( x + 5 )

= x^2 + 5x - 2x + 10

= x^2 + 3x + 10

b ) 3x + 3y +ax + ay 

= x( 3 + a ) + y( 3 + a )

= ( 3 + a )( x + y ) 

c ) ( x^2 + 2xy ) : ( x + 2y )

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

= x  : 1

= x 

d ) ( x - 2 )( x + 2 ) + ( x + 1 )^2 - 2x^2 = 0 

       x^2 + 2x - 2x - 4 + x^2 + x + x + 1 - 2x^2 = 0

       x^2 - 4 + x^2 + 2x + 1 - 2x^2 = 0

        2x^2 + 2x - 4 + 1 - 2x^2 = 0 

       2x - 3 = 0

       2x = 0 + 3

       2x = 3

        x = 3 : 2

        x = 3/2 

a)   \(\left(x-2\right)\left(x+5\right)\)

\(=x^2+5x-2x-10\)

\(=x^2+3x-10\)

b)  \(3x+3y+ax+ay\)

\(=3\left(x+y\right)+a\left(x+y\right)\)

\(=\left(x+y\right)\left(3+a\right)\)

c)   \(\left(x^2+2xy\right):\left(x+2y\right)\)

\(=\left[x\left(x+2y\right)\right]:\left(x+2y\right)\)

\(=x\)

d)  \(\left(x-2\right)\left(x+2\right)+\left(x+1\right)^2-2x^2=0\)

\(\Leftrightarrow\)\(x^2-4+x^2+2x+1-2x^2=0\)

\(\Leftrightarrow\)\(2x-3=0\)

\(\Leftrightarrow\)\(2x=3\)

\(\Leftrightarrow\)\(x=\frac{3}{2}\)

Vậy....

(x-2)(x+5)=x^2+5x-2x-10=x^2+3x-10

3x+3y+2x+2y=3(x+y)+2(x+y)=5(x+y)

(x^2+2xy):(x+2y)=x(x+2y):(x+2y)=x

x^2-4+x^2+2x+1-2x^2=0 khi 2x-3=0 khi x=3/2

câu 1:

a,x²+2x-4z²+1

=x²+2x.1+1²-(2z)²

=(x+1)²-(2z)²

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

bạn nên dùng hằng đẳng thức đã học

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

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

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

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

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

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

c)