Question

rút gọn:

\(\dfrac{x^3+y^3+z^3-3xyz}{x^2+y^2+z^2-xy-yz-xz}\)

Answer 1

Đặt \(A=x^3+y^3+z^3-3xyz\)

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

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)\cdot z+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

Đặt \(B=x^2+y^2+z^2-xy-yz-xz\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)}{x^2+y^2+z^2-xy-yz-xz}=x+y+z\)