cần gấp

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

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

\(=8x^3-4x^2-2x^2+x+2x-1=\left(2x-1\right)\left(4x^2-x+1\right)\)

= (x³+3x²+3x+1)-(4y)³

=(x+1)³-(4y)³

=(x+1-4y)[(x+1)²+(x+1).4y+16y² ]

=(x+1-4y)[(x²+2x+1)+(4xy+4y)+16y²]

\(x^6-64x^{12}=\left(x^3\right)^2-\left(8x^6\right)^2=\left(x^3-8x^6\right)\left(x^3+8x^6\right).\)

\(=x^6\left(1-8x^3\right)\left(1+8x^3\right)=x^6\left(1-2x\right)\left(1+2x+4x^2\right)\left(1+2x\right)\left(1-2x+4x^2\right)\)

(2x-1)^3

Phân tích thành nhân tử

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

Có:\(x^4+64y^4\)

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

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

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

Linz

= 64y⁴ + 32xy³ + 8y²x²  - 32xy³  -16x²y² -  4x³y + 8x²y² +4x³y +x⁴

= 8y² ( 8y² + 4xy + x² ) - 4xy ( 8y² + 4xy + x² ) + x²  ( 8y² + 4xy + x² )

= ( 8y² - 4xy + x² ) ( 8y² + 4xy + x² )

a, 8x^3 - 1/8

 = (2x)^3 - (1/2)^3

= ( 2x - 1/2) ( 4x^2 + x + 1/4)

b, 1/25.x^2 - 64y^2 = (1/5x)^2 - (8y)^2 = ( 1/5x - 8y)(1/5x+8y)

Đúng xho mình nha ) 

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,\(0,04x^2-64y^2=\left(0,2x\right)^2-\left(8y\right)^2=\left(0,2x-8y\right)\left(0,2x+8y\right)\)

b,\(-x^3+9x^2-27x+27=-x^3+3x^2+6x^2-18x-9x+27\)

\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)=\left(-x^2+6x-9\right)\left(x-3\right)\)

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

Nhớ tick mình nha bạn,cảm ơn nhiều nha.

a) 15x²-5x³=5x²(3-x)

a: \(15x^2-5x^3=5x^2\left(3-x\right)\)

b: \(8x^3-y^3+4x^2y-2xy^2\)

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

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

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

c: Ta có: \(x^8+64y^4\)

\(=x^8+16x^4y^2+64y^4-16x^4y^2\)

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

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