Question

Phân tích các đa thức sau thành nhân tử:

a, x²+2x+1

b, a³-b³+c³+3abc

c, a³-b³-c³-3abc

 

 

Answer 1

a) \(x^2+2x+1=x^2+x+x+1=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x+1\right)=\left(x+1\right)^2\)    *Câu này có thể áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\)  cho nhanh*

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

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

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

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

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

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

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

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

 

 

 

 

 

Answer 2

a,(x+1)²

b,(a+c-b).{(a+c)^2+(a+c)b+b^2-3ac}

c,(a-c-b).{(a-c)^2+(a-c)b+b^2+3ac}