4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

3n(m - 3) + 5m(m - 3)

= (3n + 5m)(m - 3)

2a(x - y) - (y - x)

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

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

a, 4a^2b^3 - 6a^3b^2 = 2a^2b^2(2b - 3a)

b, 5(a + b) +x( a + b ) = ( 5 + x )( a + b )

c, (a - b)^2 - ( b - a ) = ( a - b )^2 + ( a - b ) = (a - b) ( a - b + 1)

a: =(5a-a+b)(5a+a-b)

=(4a+b)(5a-b)

b: =(2a-a-b)(2a+a+b)

=(a-b)(3a+b)

c: =(7a-2a+b)(7a+2a-b)

=(5a+b)(9a-b)

d: =(6a-3a+2b)(6a+3a-2b)

=(3a+2b)(9a-2b)

e: =(9a-5a+3b)(9a+5a-3b)

=(4a+3b)(14a-3b)

Lời giải:

$25a^2-(a-b)^2=(5a)^2-(a-b)^2=[5a-(a-b)][5a+(a-b)]=(4a+b)(6a-b)$

$4a^2-(a+b)^2=(2a)^2-(a+b)^2=[2a-(a+b)][2a+(a+b)]=(a-b)(3a+b)$

$49a^2-(2a-b)^2=(7a)^2-(2a-b)^2=[7a-(2a-b)][7a+(2a-b)]=(5a+b)(9a-b)$

$36a^2-(3a-2b)^2=(6a)^2-(3a-2b)^2=[6a-(3a-2b)][6a+(3a-2b)]$

$=(3a+2b)(9a-2b)$

$81a^2-(5a-3b)^2=(9a)^2-(5a-3b)^2=[9a-(5a-3b)][9a+(5a-3b)]$

$=(4a+3b)(14a-3b)$

1) a^2 + b^2 + 2a - 2b - 2ab = (a^2 - 2ab + b^2) + (2a-2b) = (a-b)^2 + 2(a-b) = (a-b)(a-b+2)

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

3) a^3+6a^2+12a+8= (a^3+8)+(6a^2+12a)= (a+2)(a^2-2a+4)+6a(a+2)=(a+2)(a^2-2a+4+6a)=(a+2)(a^2+4a+4)=(a+2)(a+2)^2=(a+2)^3

\(a.25^2-4a^2+12ab-9b^2\\ =25^2-\left(4a^2+12ab-9b^2\right)\\ =25^2-\left(2a-3b\right)^2\\ =\left(25-2a+3b\right)\left(25+2a-3b\right)\\ b.x^3+x^2y-xy^2-y^3\\ =x^2\left(x+y\right)-y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-y^2\right)\\ =\left(x+y\right)\left(x+y\right)\left(x-y\right)\\ =\left(x+y\right)^2\left(x-y\right)\)

a: Ta có: \(25x^2-4a^2+12ab-9b^2\)

\(=25x^2-\left(2a-3b\right)^2\)

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

b: Ta có: \(x^3+x^2y-xy^2-y^3\)

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

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

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

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

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

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

                                          \(=\left(2a-1\right)^2-\left(2b\right)^2\)

                                            \(=\left(2a-1-2b\right)\left(2a-1+2b\right)\)

\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1

Ta có lược đồ sau :

Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)

a²-b²-4a+4b

=(a-b)(a+b)-4(a-b)

=(a-b)(a+b-4)

b,

x³-3x²-3x+1

=(x+1)(x²-x+1)-3x(x+1)

=(x+1)(x²-4x+1)

c,sai đề

mình trả lời câu a,b đã mình đang bận

a, a^2-b^2-4a+4b

=(a-b)(a+b)-4(a-b)

=(a-b)(a+b-4)

b, x^3-3x^2-3x+1

=x^3 +x^2-4x^2-4x+x+1

=x(x+1)-4x(x+1)+(x+1)

=(x+1)(x-4x+1)