\(27x^3-a^3b^3\)

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

\(=\left(3x-ab\right)\left[\left(3x\right)^2+3x\cdot ab+\left(ab\right)^2\right]\)

\(=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)

27x³ - a³b³

= (3x)³ - (ab)³

= (3x - ab)(9x² + 3xab + a²b²)

Chọn C

Chọn C

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

= -2 (a^2 -2a)(1-x)

= -2a(a-2)(1-x)

= 2a(a-2)(x-1)

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

\(=-a\cdot2a\left(x-1\right)-2\cdot2a\left(x-1\right)\)

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

Bài 2:

1)  \(x^2-4x+4=\left(x-2\right)^2\)

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

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

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

5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

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

Bài 2:

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

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

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

10) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a\cdot2b=4ab\)

11) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

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

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

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

1:

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

2: 9x^3y^2+3x^2y^2=3x^2y^2(3x+1)

3: x^3+2x^2+3x=x(x^2+2x+3)

4: 2x^2-4x=2x(x-2)

5: 3x-6y=3(x-2y)

6: x^2-3x=x(x-3)

7: 6x^2y+4xy^2+2xy

=2xy(3x+2y+1)

8: 5x^2(x-2y)-15x(x-2y)

=(x-2y)(5x^2-15x)

=5x(x-3)(x-2y)

9: =3(x-y)+5y(x-y)

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

10: =(x-1)(3x+5)

11: =2(2x-1)-3(2x-1)

=-(2x-1)

Kết quả

a) 2x + 1.           b) x – 3.         c) – ( ab   –   2 c ) 2 .

Ưm 

đề rồi sai bạn

Vui lòng sửa đề , hok tốt!

\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)

\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)

\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)

 Trả lời:

7xy⁵(x - 1) - 3x²y⁴(1 - x) + 5xy³(x - 1)

= 7xy⁵(x - 1) + 3x²y⁴(x - 1) + 5xy³(x - 1)

= (7xy⁵ + 3x²y⁴ + 5xy³)(x - 1)

= xy(7y⁴ + 3xy³ + 5y²)(x - 1)

a 3 b 3 = a 3 3 . b 3 3 = a b 3

Vế trái bằng vế phải nên đẳng thức được chứng minh.