Question

cho x/(y+z) + y/(z+x) + z(x+y)=1. tính GTBT x^2/(y+z) + y^2/(z+x) + z^2/(x+y).

huhu các bn giúp mk vs nek

Answer 1

\(\frac{x}{y+z}=1-\left(\frac{y}{z+x}+\frac{z}{x+y}\right)\)

\(=1-\frac{xy+y^2+xz+z^2}{\left(x+z\right)\left(x+y\right)}\) \(=\frac{x^2+xy+xz+yz-xy-y^2-xz-z^2}{\left(x+z\right)\left(x+y\right)}\)

\(=\frac{x^2+yz-y^2-z^2}{\left(x+y\right)\left(x+z\right)}=\frac{\left(x^2+yz-y^2-z^2\right)\left(y+z\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

\(=\frac{x^2y+x^2z-y^3-z^3}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\)

\(\Rightarrow\frac{x^2}{y+z}=\frac{x^3y+x^3z-xy^3-xz^3}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\)

+ CM tương tự rồi công vế theo vế ta đc

BT = 0