D) 64x^3-1/8y^3

= (4x)^3 + (1/2y)^3

= ( 4x + 1/2y ) [ (4x)^2 - 4x.1/2y + (1/2y)^2 ]

E) 125x^6-27y^9

( câu này mik chưa rõ nên vx chưa tek giải cho bn )

HOk tốt nhé

a.x^3 + 8

= x^3 + 2^3

= ( x+2) ( x^2 - x.2+2^2)

b/27-8y^3

3^3 - (2y)^3

= ( 3 - (2y))(3^2 + 3.2y + (2y)^2)

c/ y^6 + 1

= (y^3)^3 + 1 ^3

= (y^3 + 1)((y^3)^2 - y^3.1 + 1^2

Cảm ơn bn nhìu ><

1/ x^2 +4xy +4y^2 = (x +2y)^2

2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3

3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3

4/ x^3 - 6x^2y+12xy^2 -8y^3= (x-2y)^3

1) x² + 4xy + 4y² = ( x + 2y )²

2) - x³ + 9x² - 27x + 27 = ( 3 - x )²

3) 8x⁶ + 36x⁴y + 54x²y² + 27y³ = ( 2x² + 3y )³

4) x³ - 6x²y + 12xy² - 8y³ = ( x - 2y )³

5) x² + 4y² +1 - 4xy - 2x + 4y = ( x² - 2y - 1 )²

6) x² + y² + 4 + 2xy + 4x + 4y = ( x + y + 2 )²

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

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

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

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

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

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

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

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

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

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

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

\(a^6-8b^3=\left(a^2\right)^3-\left(2b\right)^3=\left(a^2-2b\right)\left(a^4+2a^2b+4b^2\right)\)

\(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)

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

1. x³ + 8 = (x + 2 )(x² - x + 1)

2. 27 - 8y³ = ( 3 - 2y ) ( 9 + 6y + 4y² )

3. y⁶ + 1 = (y²)³ + 1 = ( y² + 1) ( y⁴ - y² +1 )

4.64x³ - \(\dfrac{1}{8}\)y³ = ( 4x - \(\dfrac{1}{2}\)y ) ( 16x² + 2xy + \(\dfrac{1}{4}\)y²)

5. 125x⁶ - 27y⁹ = (5x²)³ - (3y³)³

= ( 5x² - 3y³)(25x⁴ +15x²y³ + 9y⁶)

b)B=27y^3-27y^2x+9yx^2-x^3 
= 27 . (1/3x)^3 - 27.(1/3x)².x + 9.1/3.x.x^2 - x^3 
= x^3 - 3x^3 + 3x^3 - x^3 
= 0

d) D=50y^2+x(x-2y)+14y(x-y) 

=50y^2 +x^2 -2xy +14xy -14y^2 

=36y^2 +x^2 +12xy 

=(6y + x)^2 

=81 

Lời giải:

a.

$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$

b.

$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$

c.

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

d.

$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$

$=(x+y)(x^2-4xy+7y^2)$

a) \(64x^2-24y^2\)

\(=8\left(8x^2-3y^2\right)\)

b) \(64x^3-27y^3\)

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

\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)

c) \(x^4-2x^3+x^2\)

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

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

d) \(\left(x-y\right)^3+8y^3\)

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

\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)

im chưa học\

Ta có : x² + 3x 

= x² + \(2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)

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

Ta có : (2x - 3)(4x² - 6x + 9) 

= (2x -3)(2x + 3)²

= (4x² - 9)(2x + 3)