4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

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)

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

Ta có lược đồ sau :

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

dấu ^ là mũ nha mn

e) Ta có: \(a^3-a^2-a+1\)

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

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

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

f) Ta có: \(x^3-2xy-x^2y+2y^2\)

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

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

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

b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

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

d) Đề sai ko ???

e) \(a^3-a^2-a+1=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)=\left(a-1\right)^2\left(a+1\right)\)

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

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

\(b,=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

\(c,=-\left(x^2-3x^2+3x-1\right)=-\left(x-1\right)^3\)

\(d,=2\left(x^2+2xy+y^2-4z^2\right)=2\left(\left(x+y\right)^2-4z^2\right)=2\left(x+y-2z\right)\left(x+y+2z\right)\)

\(e,=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)\)

\(f,=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)

a) (a+b+c)^2 + (a+b-c)^2 - 4c^2

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

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

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

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

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

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

b) 4a^2b^2 - (a^2+b^2-c^2)^2

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

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

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

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

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

\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)

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

\(=a^3\left(c-b\right)+\left(c-b\right)\left(bc^2+b^2c\right)-\left(c-b\right)\left(ac^2+abc+ab^2\right)\)

\(=\left(c-b\right)\left(a^3+bc^2+b^2c-ac^2-abc-ab^2\right)\)

a) (a+b+c)^2 + (a+b-c)^2 - 4c^2

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

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

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

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

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

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

b) 4a^2b^2 - (a^2+b^2-c^2)^2

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

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

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

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

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

\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)

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

\(=a^3\left(c-b\right)+\left(c-b\right)\left(bc^2+b^2c\right)-\left(c-b\right)\left(ac^2+abc+ab^2\right)\)

\(=\left(c-b\right)\left(a^3+bc^2+b^2c-ac^2-abc-ab^2\right)\)

a)\(a^2+6a+8-b^2-2b=\left(a+3\right)^2-\left(b+1\right)^2=\left(a+3+b+1\right)\left(a+3-b-1\right)\)

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

b)\(a^2+6ax+8x^2-b^2-2bx\)

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

\(=\left(a+3x-b-x\right)\left(a+3x+b+x\right)=\left(a-b+2x\right)\left(a+b+4x\right)\)