Question

[Trang]

Phân tích đa thức thành nhân tử:

1) \(a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2\)

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

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

Answer 1

[tthnew]

2) Để sau đi (em chưa nghĩ ra)

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

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

Đặt x - y = a; y - z = b => z - x = -(a+b)

\(A=\left(x+y\right)^2a+\left(y+z\right)^2b-\left(z+x\right)^2a-\left(z+x\right)^2b\)

\(=a\left[\left(x+y\right)^2-\left(z+x\right)^2\right]+b\left[\left(y+z\right)^2-\left(z+x\right)^2\right]\)

\(=\left(x-y\right)\left(x+y-z-x\right)\left(x+y+z+x\right)+\left(y-z\right)\left(y+z-z-x\right)\left(y+z+z+x\right)\)

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

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

Em tính sai sót chỗ nào thì thông cảm cho em ạ :>

Answer 2

[hattori heiji]

1) $a^4+b^4+c^4-2a^2{b}^2-2a^2{c}^2-2b^2{c}^2$

=2(a⁴+b⁴+c⁴-4a²b²-4a²c²-4b²c²)

=2a⁴+2b⁴+2c⁴-4a²b²-4a²c²-4b²c²

=(a4-2a2b2+b4)+(a4-2a2c2+c4)+(b4-2b2c2+c4