Question

Phân tích đa thức thành nhân tử:

a, \(x^3-3x^2+3x-1-y^3\)

b. \(8x^2+10x-3\)

c, \(x^2-8xy+15y^2+2x-4y-3\)

Answer 1

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

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

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

\(=\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)

....

Answer 2

\(8x^2+10x-3\)

\(=8x^2+12x-2x-3\)

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

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

\(x^3-3x^2+3x-1-y^3\)

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

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

ps: lớp 7, ko chắc 

Answer 3

eeeeiiiiiiiii

câu a ấy quên cái ngoặc nhoa :<

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