Question

[Vũ Thành Minh]

b1: phân tích cá đa thức sau thành nhân tử 

bt1

a)x^2-y^2+8x-8y

b)4x^2+4xy+y^2-4x-2y

c)x^3+y^3+4x+4y

d)x^3-3x^2y+3xy^2-y^3+x^2-y^2

e)x^2+3x+2

f)x^2-7x+6

h)x^2-3x-2

Answer 1

[Nguyễn Đức Trí]

a) \(x^2-y^2+8x-8y=\left(x-y\right)\left(x+y\right)+8\left(x-y\right)=\left(x-y\right)\left(x+y+8\right)\)

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

c)\(x^3+y^3+4x+4y=\left(x+y\right)\left(x^2-xy+y^2\right)+4\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2+4\right)\)

d) \(...=\left(x-y\right)^3+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left[\left(x-y\right)^2+x+y\right]\)

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

f) \(x^2-7x+6=x^2-x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)

Answer 2

[Đào Trí Bình]

sao mình lại ko hiểu nhỉ