c) a3 – b3 + 2b – 2a = (a – b)(a2 + ab + b2) – 2(a – b)

=(a – b)( a2 + ab + b2 – 2)

a(b³ - c³) + b(c³ - a³) + c(a³ - b³)

= a(b³ - c³ ) + b( c³ - b³ + b³ - a³) + c(a³ - b³)

= a(b³ - c³) + b(c³ - b³) + b(b³ - a³) + c(a³ - b³)

= a(b³ - c³) - b(b³ - c³) - [b(a³ - b³) - c(a³- b³)]

= (b³ - c³)(a - b) - (a³- b³)(b - c)

= (b - c)(b² + bc + c²)(a - b) - (a - b)(a² + ab + b²)(b - c)

= (b - c)(a - b)(b² + bc + c² - a² + ab - b²)

= (b - c)(a - b) [ (c²  - a²) + (bc - ab) ]

= (b - c)(a - b) [ (c - a)(c + a) + b(c - a) ]

= (b - c)(a -b) [ (c - a)(c + a + b) ]

= (a- b)(b - c)(c - a)(a + b + c)

Ta có: \(a^2-4b^2-a^3+4a^2b-4ab^2\)

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

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

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

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

a: \(10x^3-90x^5\)

\(=10x^3\cdot1-10x^3\cdot9x^2\)

\(=10x^3\left(1-9x^2\right)=10x^3\left(1-3x\right)\left(1+3x\right)\)

b: \(3a^2-6a^2b+5a-10ab\)

=3a^2(1-2b)+5a(1-2b)

=(1-2b)(3a^2+5a)

=a(3a+5)(1-2b)

c: \(7a^3-28a^2+28a\)

\(=7a\left(a^2-4a+4\right)\)

\(=7a\left(a-2\right)^2\)

d: \(45a^3-30a^2+5a-500\)

\(=5\left(9a^3-6a^2+5a-100\right)\)

e:Sửa đề: \(a^2b-49b+14b^2-b^3\)

\(=b\left(a^2-49+14b-b^2\right)\)

\(=b\left\lbrack a^2-\left(b-7\right)^2\right\rbrack\)

=b(a-b+7)(a+b-7)

f: \(x^4+4\)

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

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

a) 10x³-90x⁵=10x³(1-9x²)=10x³(1-3x)(1+3x)

b)3a²-6a²b+5a-10ab=(3a²-6a²b)+(5a-10ab)=3a²(1-2b)+5a(1-b)=(3a²+5a)(1-2b)=a(3a+5)(1-2b)

c) 7a³-28a²+28a=7a(a²-4a+4)

d) 45a³-30a²+5a-500=5(9a³-6a²+a-100)

Ta có

a 4 + a 3 + a 3 b + a 2 b = a 4 + a 3 + a 3 b + a 2 b = a 3 a + 1 + a 2 b a + 1 = a + 1 a 3 + a 2 b = a + 1 a 2 a + b = a 2 a + b a + 1

Đáp án cần chọn là: A

`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`

\(2x^2+4ax+x+2a\)

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

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