Question

x/(y-z)+y/(z-x)+z/(x-y)=0. Tính x/(y-z)^2+y/(z-x)^2+z/(x-y)^2

Answer 1

\(\dfrac{x}{y-z}+\dfrac{y}{z-x}+\dfrac{z}{x-y}=0\)

\(\Rightarrow\dfrac{x}{y-z}=-\dfrac{y}{z-x}-\dfrac{z}{x-y}=\dfrac{y}{x-z}-\dfrac{z}{x-y}\\\Leftrightarrow\dfrac{x}{y-z}=\dfrac{y(x-y)-z(x-z)}{(x-z)(x-y)}=\dfrac{xy-y^{2}-xz+z^{2}}{(x-z)(x-y)}\\\Leftrightarrow \dfrac{x}{(y-z)^{2}}=\dfrac{xy-y^{2}-xz+z^{2}}{(x-z)(x-y)(y-z)} \ \ \ \ (1)\)

Hoán vị vòng quanh x→y→z→x ta đc:

\(\dfrac{y}{(z-x)^{2}}=\dfrac{yz-z^{2}-yx+x^{2}}{(y-x)(y-z)(z-x)} \ \ \ \ (2)\\\dfrac{z}{(x-y)^{2}}=\dfrac{zx-x^{2}-zy+y^{2}}{(z-y)(z-x)(x-y)} \ \ \ \ (3)\)

Cộng vế vs vế của (1),(2) và (3) ta đc \(\dfrac{x}{(y-z)^{2}}+\dfrac{y}{(z-x)^{2}}+\dfrac{z}{(x-y)^{2}}=0\)