Question

Given that x/y+z+y/z+x+z/x+y=1

Evaluate A=X^2/y+z+y^2/Z+x+z^2/x+y

Answer 1

\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\\ =>\dfrac{x}{y+z}=1-\dfrac{y}{z+x}-\dfrac{z}{x+y}\\ =>\dfrac{x}{y+z}=\dfrac{(z+x)(x+y)-y(x+y)-z(z+x)}{(z+x)(x+y)}\\ =>\dfrac{x}{y+z}=\dfrac{xz+yz+x^{2}+xy-xy-y^{2}-z^{2}-xz}{(z+x)(x+y)}\\ =>\dfrac{x}{y+z}=\dfrac{x^{2}-y^{2}-z^{2}+yz}{(z+x)(x+y)}\\ =>\dfrac{x^{2}}{y+z}=\dfrac{x^{3}-xy^{2}-xz^{2}+xyz}{(z+x)(x+y)} \ \ \ \ (1)\\ =>\dfrac{y^{2}}{z+x}=\dfrac{y^{3}-yz^{2}-yx^{2}+xyz}{(x+y)(y+z)} \ \ \ \ (2)\\ =>\dfrac{z^{2}}{x+y}=\dfrac{z^{3}-zx^{2}-zy^{2}+xyz}{(y+z)(z+x)} \ \ \ \ (3)\)

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