1. 5x² - 4x

= x(5x - 4)

2. 8x²(x - 3y) - 12x(x - 3y)

= (8x² - 12x)(x - 3y)

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

3. 3(x - y) - 5x(y - x)

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

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

1- hay câu 1 vậy bn

=-(y-2x-1)(y²-4xy+y+4x²-2x+1)

c: \(=\left(5x-y\right)\left(5x+y\right)\)

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

a) x(4y-10x)

b)3(x+2y)+(x+1)

c)(5x-y)(5x+y)

d)5x(y-z)²

e)(x-3)(x-2)

f)(2x+y)³

\(2x-1^3+8\)

\(=2x-9\)

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

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

_________

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

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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

_______________

\(8x^3-12x^2+6x-2\)

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

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

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

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

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

________

\(9x^3-12x^2+6x-1\)

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

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

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

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

b: 8x^3-12x^2+6x-1

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

=(2x-1)^3

c: =(8x^3-12x^2+6x-1)-1

=(2x-1)^3-1

=(2x-1-1)[(2x-1)^2+2x-1+1]

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

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

8x³ - 12x² + 6x - 1

= (2x)³ - 3.(2x)².1 + 3.2x.1 - 1³

= (2x - 1)³

--------------------

8x³ - 12x² + 6x - 2

= 8x³ - 12x² + 6x - 1 - 1

= (2x)³ - 3.(2x)².1 + 3.(2x).1 - 1³ - 1³

= (2x - 1)³ - 1³

= (2x - 1 - 1)[(2x - 1)² + (2x - 1).1 + 1]

= (2x - 2)(4x² - 4x + 1 + 2x - 1 + 1)

= 2(x - 1)(4x² - 2x + 1)

--------------------

9x³ - 12x² + 6x - 1

= x³ + 8x³ - 12x² + 6x - 1

= x³ + (2x)³ - 3.(2x)² + 3.2x.1² - 1³

= x³ + (2x - 1)³

= (x + 2x - 1)[x² - x.(2x - 1) + (2x - 1)²]

= (3x - 1)(x² - 2x² + x + 4x² - 4x + 1)

= (3x - 1)(3x² - 3x + 1)

8x^3-12x^2+6x-1

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

=(2x-1)^3

`= (2x)^3 - 3*4x^2*1 + 3*2x*1-1^3`

`= (2x - 1)^3`

a,  ( 2x - 1)^2 - (4x + 2) ^2 = ( 2x - 1 - 4x- 2) ( 2x - 1 + 4x + 2) = (-2x-3)(6x+1) = - (2x+3)(6x+1)

b, 8x^3 + 12x^2y + 6xy^2 + y^3 

= (2x)^2 + 3.(2x)^2 . y + 3.2x.y^2 + y^3 

= (2x + y)^3 

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

a) Ta có: \(x^3+12x^2+48x+64\)

\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3\)

\(=\left(x+4\right)^3\)

b) Ta có: \(x^3-12x^2+48x-64\)

\(=x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3\)

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

c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

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

d)Sửa đề: \(x^3-3x^2+3x-1\)

Ta có: \(x^3-3x^2+3x-1\)

\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)

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

e) Ta có: \(8-12x+6x^2-x^3\)

\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)

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

f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)

\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)

\(=\left(\frac{1}{3}-3y\right)^3\)