Question

[Hoàng Quý]

Phân tích đa thứ sau thành nhân tư

a) x^6+x^4+x^2.y^2+y^4-y^6

b) 9(a+b)^2+16a-4b-(a-b)^2-24

c) (2a+b)^3+6a+3b-4

Answer 1

[Huy Thắng Nguyễn]

a) \(x^6+x^4+x^2y^2+y^4-y^6\)

\(=x^6-y^6+x^4+x^2y^2+y^4\)

\(=\left[\left(x^3\right)^2-\left(y^3\right)^2\right]+x^4+2x^2y^2+y^4-x^2y^2\)

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

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

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

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

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

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

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

Answer 2

[Huy Thắng Nguyễn]

c) \(\left(2a+b\right)^3+6a+3b-4\)

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

Đặt 2a + b = t.

Ta có: \(t^3+3t-4\)

\(=t^3-t^2+t^2-t+4t-4\)

\(=t^2\left(t-1\right)+t\left(t-1\right)+4\left(t-1\right)\)

\(=\left(t-1\right)\left(t^2+t+4\right)\)

Thay t = 2a + b vào biểu thức:

\(\left(t-1\right)\left(t^2+t+4\right)=\left(2a+b-1\right)\left(4a^2+4ab+b^2+2a+b+4\right)\)