Question

Phân tích đa thức thành nhân tử

x(y²-z²)+z(x²-y²)+y(z²-x²)

[4abcd+(a²+b²).(c²+d²)]-4[cd(a²+b²)+ab(c²+d²)]²

Answer 1

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

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

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

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

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

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