Question

[nguyễn bảo thuận]

phân tích đa thức thành nhân tử

1) (x+y)³-x³-y³                          2) (a²+b²-5)²-4(ab+2)²

3)bc(b+c)+ac(c-a)-ab(a+b)       4)(x-y)³+(y-z)³+(z-x)³giúp mik vs mik gắp lắm r

Answer 1

[Đoàn Đức Hà]

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

2) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

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

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

4) \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

Ta có hằng đẳng thức: 

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

Chứng minh: 

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

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

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

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

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

Áp dụng ta được: 

\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3-3\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\)

\(\Leftrightarrow\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\).