Question

Phân tích đa thức thành nhân tử : ( phương pháp đổi biến )

a ) \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)

b ) \(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)

c ) \(2\left(x^4+y^4+z^4\right)-\left(x^2+y^2+z^2\right)^2-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)

Answer 1

1.

$A=(x+a)(x+2a)(x+3a)(x+4a)+a^4=[(x+a)(x+4a)][(x+2a)(x+3a)]+a^4$

$=(x^2+5ax+4a^2)(x^2+5ax+6a^2)+a^4$

Đặt $x^2+5ax+4a^2=t$ thì $A=t(t+2a^2)+a^4=t^2+2ta^2+a^4=(t+a^2)^2$

$=(x^2+5ax+4a^2+a^2)^2=(x^2+5ax+5a^2)^2$

Answer 2

2.

$B=(x^2+y^2+z^2)(x+y+z)^2+(xy+yz+xz)^2$

$=(x^2+y^2+z^2)[(x^2+y^2+z^2)+2(xy+yz+xz)]+(xy+yz+xz)^2$

Đặt $x^2+y^2+z^2=a; xy+yz+xz=b$ thì:

$B=a(a+2b)+b^2=a^2+2ab+b^2=(a+b)^2$

$=(x^2+y^2+z^2+xy+yz+xz)^2$