4acx + 4bcx + 4ax + 4bx

= 4cx ( a+ b) + 4x ( a+b)

= 4x ( c +1 )(a +b)

Phân tích đa thức thành nhân tử:

\(4acx+4bcx+4ax+4bx\)

\(=\left(4acx+4ax\right)+\left(4bcx+4bx\right)\)

\(=4ax\left(c+1\right)+4bx\left(c+1\right)\)

\(=\left(4ax+4bx\right)\left(c+1\right)\)

\(=4x\left(a+b\right)\left(c+1\right)\)

a) x² ( x+ 2y) -x -2y

= x² ( x+ 2y) -(x+2y)

= (x²-1)(x+2y)

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

b)3x²- 3y² -2 (x-y)²

= 3(x²-y²) -2 (x-y)²

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

\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\\ =\left(x-y\right)\left(3x+3y-2x+2y\right)\\ =\left(x-y\right)\left(x+5y\right)\)

c) x²- 2x-4y² - 4y

= (x²-4y²)-(2x+4y)

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

d) x³ - 4x² - 9x +36

= (x³+3x²)-(7x²+21x)+(12x+36)

= x²(x+3)-7x(x+3)+12(x+3)

=(x²-7x+12)(x+3)

\(=\left[\left(x^2-3x\right)-\left(4x-12\right)\right]\left(x+3\right)\\ =\left[x\left(x-3\right)-4\left(x-3\right)\right]\left(x+3\right)=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

a) = x² ( x+ 2y) -(x+2y)

= (x²-1)(x+2y)

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

b)= 3(x²-y²) -2 (x-y)²

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

=(x−y)[3(x+y)−2(x−y)]

=(x−y)(3x+3y−2x+2y)

=(x−y)(x+5y)

=(x−y)[3(x+y)−2(x−y)]

=(x−y)(3x+3y−2x+2y)

=(x−y)(x+5y)

c)= (x²-4y²)-(2x+4y)

=(x−2y)(x+2y)−2(x+2y)

=(x+2y)(x−2y−2)

=(x−2y)(x+2y)−2(x+2y)

=(x+2y)(x−2y−2)

d)= (x³+3x²)-(7x²+21x)+(12x+36)

= x²(x+3)-7x(x+3)+12(x+3)

=(x²-7x+12)(x+3)

=[(x2−3x)−(4x−12)](x+3)

=[x(x−3)−4(x−3)](x+3)

=(x−4)(x−3)(x+3)

a: \(x^2\left(x+2y\right)-x-2y\)

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

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

b: \(3x^2-3y^2-2\left(x-y\right)^2\)

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

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

\(=\left(x-y\right)\left(x+5y\right)\)

c: Ta có: \(x^2-2x-4y^2-4y\)

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

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

d: Ta có: \(x^3-4x^2-9x+36\)

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

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

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

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

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

b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

c) Đổi đề: \(a^2x+a^2y-7x-7y\)

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

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

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

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

g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)

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

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

b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

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

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

g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)

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

-- là giề

a(b^3-c^3) +b(c^3-a^3)+c(a^3-b^3)

=> a(b-c)(b^2+bc+c^2)+bc^3-ba^3+ca^3-cb^3

=>a(b-c)(b^2+bc+c^2)-(cb^3-bc^3)-(ba^3-ca^3)

=>a(b-c)(b^2+bc+c^2)-bc(b-c)(b+c)-a^3(b-c)

=>(b-c)(ab^2+abc+ac^2-cb^2-bc^2-a^3)

=>(b-c)(

a(b³-c³)+b(c³-a³)+c(a³-b³)

=a.b³+b.c³+c.a³-a.c³-b.a³-c.b³

=(a.b³-b.a³)+(b.c³-c.b³)+(c.a³-c³.a)

=(ab.b²-ba.a²)+(bc.c²-bc.b²)+(ca.a²-ca.c²)

=ab(a²-b²)+bc(c²-b²)+ca(c²-a²)

có hằng đẳng thức nào ko

Bạn đưa ra 1 ví dụ đi rồi mình giảng

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}