Y = a^3 - b^3

Y = ( a - b ) ( a^2 + a . b + b^2 )

\(\dfrac{8x^3y^2-6x^2y^3}{-2xy}=\dfrac{8x^3y^2}{-2xy}+\dfrac{6x^2y^3}{2xy}=-4x^2y+3xy^2\)

⇒ Chọn A.

Câu 1: C

Câu 2: B

14.C

15.D

Câu 14: C

Câu 15: D

1. 

a)      \({\left( {x + 3} \right)^3} = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {x^3} + 9{x^2} + 27x + 27\)

b)      \({\left( {x + 2y} \right)^3} = {x^3} + 3.{x^2}.2y + 3.x.{\left( {2y} \right)^2} + {\left( {3y} \right)^3} = {x^3} + 6{x^2}y + 12x{y^2} + 27{y^3}\)

2. 

\(\begin{array}{l}{\left( {2x + y} \right)^3} - 8{x^3} - {y^3} = {\left( {2x} \right)^3} + 3.{\left( {2x} \right)^2}.y + 3.2x.{y^2} + {y^3} - 8{x^3} - {y^3}\\ = 8{x^3} + 12{x^2}y + 6x{y^2} + {y^3} - 8{x^3} - {y^3}\\ = \left( {8{x^3} - 8{x^3}} \right) + 12{x^2}y + 6x{y^2} + \left( {{y^3} - {y^3}} \right)\\ = 12{x^2}y + 6x{y^2}\end{array}\)

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

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

\(\left(2y-3\right)^3=8y^3-36y^2+54y-27\)

a: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=6x^2y+2y^3\)

a) (x² + 2)²

= (x²)² + 2.x².2 + 2²

= x⁴ + 4x² + 4

b) (x + y + z)²

= [(x + y) + z]²

= (x + y)² + 2(x + y).z + z²

= x² + 2xy + y² + 2xz + 2yz + z²

= x² + y² + z² + 2xy + 2xz + 2yz

sos

a) (x - 1/2x²y)²

= x² - 2x . 1/2 x²y + (1/2x²y)²

= x² - x³y + 1/4 x⁴y²

b) (2xy² - 1)(1 + 2xy²)

= (2xy²)² - 1²

= 4x²y⁴ - 1

c) (x - y + 2)²

= (x - y)² + 2(x - y).2 + 2²

= x² - 2xy + y² + 4x - 4y + 4

= x² + y² - 2xy + 4x - 4y + 4

d) (x + 1/2)(1/2 - x)

= (1/2)² - x²

= 1/4 - x²

e) (x² - 1/3)²

= (x²)² - 2x².1/3 + (1/3)²

= x⁴ - 2/3 x² + 1/9

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

2, đề sai 

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

4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)

5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)

tương tự ... 

8, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

Câu 2 đề ko sai nha bạn.

2) x² - (\(\sqrt{y^3}\))²      ( y>0)   

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