Bài 1: 

a: \(4a^2-6b=2\left(2a^2-3b\right)\)

b: \(m^3n-2m^2n^2-mn\)

\(=mn\left(m^2-2mn-1\right)\)

Bài 1:

a) \(4a^2-6b=2\left(a^2-3b\right)\)

b) \(=mn\left(m^2-2mn-1\right)\)

Bài 2:

a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)

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

Trả lời:

\(27a^2b^2-18ab+3\)

\(=3\left(9a^2b^2-6ab+1\right)\)

\(=3\left[\left(3ab\right)^2-2.3ab.1+1^2\right]\)

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

`a, 9x^2 - 16 = (3x+4)(3x-4)`

`b, 4x^2 - 12xy + 9y^2 = (2x-3y)^2`

`c, t^3-8 = (t-2)(t^2 - 2t + 4)`

`d, 2ax^3y^3 + 2a = 2a(x^3y^3 + 1) = 2a(xy+1)(x^2y^2 - xy + 1)`

a) \(\left(9x^2-16\right)=\left(3x-4\right)\left(3x+4\right)\)

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

c) \(t^3-8=\left(t-2\right)\left(t^2+2t+4\right)\)

d) \(2ax^3y^3+2a=2a\left(x^3y^3+1\right)\)

\(a,=ab\left(a+3\right)\\ b,=\left(x-1\right)^2\\ c,=x\left[\left(x-3\right)^2-y^2\right]=x\left(x-y-3\right)\left(x+y-3\right)\)

a: =(x-1)²

\(a.6x^3-9x^2=3x^2\left(2x-3\right)\\ b.20x^2y-12x^3=4x^2\left(5y-3x\right)\)

a) \(6x^3-9x^2=3x^2\left(2x-3\right)\)

b) \(20x^2y-12x^3=4x^2\left(5y-3x\right)\)

a: \(6x^3-9x^2=3x^2\left(2x-3\right)\)

b: \(20x^2y-12x^3=4x^2\left(5y-3x\right)\)

a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

Gọi a+b =x có:  

a(x+b)³−b(x+a)³

 =a(x³+3x²b+3xb²+b³)−b(x³+3x²a+3xa²+a³)  

=ax³+3ax²b+3axb²+ab³−bx³−3bx²a−3bxa²−ba³  

=(a−b)x³+(3ax²b−3bx²a)+(3axb²−3bxa²)+ab³−ba³  

=(a−b)x³+3axb(b−a)+ab(b²−a²)  

=−x³(b−a)+3axb(b−a)+ab(b+a)(b−a)

 =−x³(b−a)+3axb(b−a)+(a²b+ab²)(b−a)

 =(b−a)(−x³+3axb+a²b+ab²)

nho lik e