1)\(8x^6-\frac{1}{125}y^3=\left(2x^2\right)^3-\left(\frac{1}{5}y\right)^3\)

Bạn tự lm tiếp.AD HĐT số (7)

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

AD HĐT số (7).Tự lm tiếp

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

AD HĐT số (7).Tự lm tiếp

4)\(x^9+1=\left(x^3\right)^3+1\)

AD HĐT số (7).Tự lm tiếp

5,\(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2\)

AD HĐT số (3).Tự lm tiếp

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

AD HĐT số (4)

7)\(x^3-15x^2+75x-125=\left(x-5\right)^3\)

AD HĐT số (5)

8)\(27a^3-54a^2b+36ab^2-8b^3\)

\(=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)

\(=\left(3a-2b\right)^3\)

AD HĐT số (5)

x⁹ + 1
= (x³)³ + 1³
= (x³ + 1)(x⁶ - x³ + 1)
= (x + 1)(x² - x + 1)(x⁶ - x³ +1)

8a³ - 12a² + 6a - 1
= (2a)³ - 3(2a)²1 + 3 . 2a . 1² - 1
= (2a - 1)³

27a³ - 54a²b + 36ab² - 8b³
= (3a)³ - 3(3a)²2b + 3 . 3a . (2b)² - (2b)³
= (3a - 2b)³

1. \(x^3+9x^2+27x+27=\left(x+3\right)^3\)

2.\(8x^6-27y^3=\left(2x\right)^3-\left(3y\right)^3=\left(2x-3y\right)\)

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

3.\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)4.câu cuối là \(8b^3\)bạn nhé !!

tik mik nhé

a: =4x^2+8x-3x-6

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

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

b: =3(3x^2-2x-1)

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

=3(x-1)(3x+1)

c: =2x^2-4x+x-2

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

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

d: =3x^2+3x-2x-2

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

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

e: =3x^2+9x+x+3

=3x(x+3)+(x+3)

=(x+3)(3x+1)

a) \(4x^2+5x-6\)

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

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

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

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

b) \(9x^2-6x-3\)

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

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

\(=3\left[3x\left(x-1\right)+\left(x-1\right)\right]\)

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

c) \(2x^2-3x-2\)

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

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

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

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

d) \(3x^2+x-2\)

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

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

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

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

e) \(3x^2+10x+3\)

\(=3x^2+9x+x+3\)

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

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

1. \(\left(x+1\right)^3-125\)

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

\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)

2. \(\left(x+4\right)^3-64\)

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

\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)

3. \(x^3-\left(y-1\right)^3\)

\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)

\(\)4. \(\left(a+b\right)^3-c^3\)

\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)

5. \(125-\left(x+2\right)^3\)

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

\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)

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

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

\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

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= (5x-25) + (5x - x²)

= 5(x-5) + x(5-x)

= 5(x-5) - x(x-5)

= (5 - x)(x - 5)

\(10x-25-x^2=-\left(x^2-2\cdot x\cdot5+5^2\right)=-\left(x-5\right)^2\)

27x⁶ + 125y⁶ = ( 3x² )³ + ( 5y² )³ = ( 3x² + 5y² )( 9x⁴ - 15x²y² + 25y⁴ )

8a⁶ - 8b⁶ = ( 2a² )³ - ( 2b² )³ = ( 2a - 2b )( 4a² + 4ab + 4b² ) = 2( a - b )4( a² + ab + b² ) = 8( a - b )( a² + ab + b² )

x⁴ + 64y⁴ = x⁴ + 16x²y² + 64y⁴ - 16x²y² 

                = ( x⁴ + 16x²y² + 64y⁴ ) - 16x²y²

                = ( x² + 8y² )² - ( 4xy )²

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

x⁴ + x³ + 2x² + x + 1 = x⁴ + x³ + x² + x² + x + 1

                                  = ( x⁴ + x³ + x² ) + ( x² + x + 1 )

                                  = x²( x² + x + 1 ) + ( x² + x + 1 )

                                  = ( x² + x + 1 )( x² + 1 )

\(27x^6+125y^6=\left(3x^2\right)^3+\left(5y^2\right)^3=\left(3x^2+5y^2\right)\left(9x^4-15x^2.y^2+25y^4\right)\)

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

                                                       \(=8\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(x^{\text{4}}+64y^4=x^4+64y^4+16x^2y^2-16x^2y^2\)

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

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

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

                Bài làm :

a) 27x⁶ + 125y⁶ 

= ( 3x² )³ + ( 5y² )³ 

= ( 3x² + 5y² )( 9x⁴ - 15x²y² + 25y⁴ )

b) 8a⁶ - 8b⁶ 

= ( 2a² )³ - ( 2b² )³ 

= ( 2a - 2b )( 4a² + 4ab + 4b² )

= 2( a - b )4( a² + ab + b² )

= 8( a - b )( a² + ab + b² )

c) x⁴ + 64y⁴ 

= x⁴ + 16x²y² + 64y⁴ - 16x²y² 

= ( x² + 8y² )² - ( 4xy )²

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

d) x⁴ + x³ + 2x² + x + 1

= x⁴ + x³ + x² + x² + x + 1

 = ( x⁴ + x³ + x² ) + ( x² + x + 1 )

 = x²( x² + x + 1 ) + ( x² + x + 1 )

 = ( x² + x + 1 )( x² + 1 )

a/ x² + 6x + 9 = (x + 3)² = (x + 3)(x + 3)

b/ 10x - 25 - x² = -x² + 10x - 25 = -(x² -10x + 25) = -(x - 5)² = -(x - 5)(x - 5)

c/ \(8x^3+\frac{1}{8}=\left(2x\right)^3+\left(\frac{1}{2}\right)^3=\left(2x+\frac{1}{2}\right)\left(4x^2-x+\frac{1}{4}\right)\)