a: Ta có: \(A=10x^2+20xy+10y^2-90\)

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

\(=10\left(x+y-3\right)\left(x+y+3\right)\)

b: Ta có: \(B=x^3y-3x^2y-4xy+12y\)

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

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

c: Ta có: \(C=125x^3-10x^2+2x-1\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)

\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)

a) 10x³-90x⁵=10x³(1-9x²)=10x³(1-3x)(1+3x)

b)3a²-6a²b+5a-10ab=(3a²-6a²b)+(5a-10ab)=3a²(1-2b)+5a(1-b)=(3a²+5a)(1-2b)=a(3a+5)(1-2b)

c) 7a³-28a²+28a=7a(a²-4a+4)

d) 45a³-30a²+5a-500=5(9a³-6a²+a-100)

a) \(x^4-y^4\)

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

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

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

b) \(x^2-3y^2\)

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

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

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

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

\(=0\cdot0\)

\(=0\)

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

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

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

\(=\left(x-5y\right)\left(5x-y\right)\)

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

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

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

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

f) \(x^3+27\)

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

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

g) \(27x^3-0,001\)

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

\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)

h) \(125x^3-1\)

\(=\left(5x\right)^3-1^3\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)

\(a,=\left(3x+\dfrac{y}{2}\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{y^2}{4}\right)\\ b,=\left(5x+3y\right)\left(25x^2+15xy+9y^2\right)\)

(2x-1)³-(2x-1)=(2x-1) [( 2x-1)² -1 ]

                      = 2x (2x -1 ) ( 2x-2)

(2x - 1)³ - (2x - 1)

= (2x - 1)[(2x - 1)² - 1]

= (2x - 1)(4x² - 4x + 1 - 1)

= (2x - 1)(4x² - 4x)

Trả lời:

( 2x - 1 )³ - ( 2x - 1 )

= ( 2x - 1 ) [ ( 2x - 1 )² - 1 ]

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

= 2x ( 2x - 1 ) ( 2x - 2 ) 

đề bài sai r bạn

Mở SGK bài Những hằng đẳng thức đáng nhớ ra

Bài cơ bản cũng ko biết làm thì nói làm gì

\(\left(2x-1\right)^3=8x^3-12x^2+6x-1\)

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

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

x³ + 2x² + 2x + 1 

= (x³ + 1) + (2x² + 2x)

= (x + 1)(x² + x + 1) + 2x(x + 1)

= (x + 1)(x² + x + 1 + 2x)

= (x + 1)(x² + 3x + 1)

 Chúc bạn học tốt

Ta có: \(x^3+2x^2+2x+1\)

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

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

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

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

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

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

#)Giải :

\(x^3-2x-4\)

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

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

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

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

\(x^4+2x^3+5x^2+4x-12\)

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

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

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

\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

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

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

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

cảm ơn nha!