Question

cho x/(y+z) + y/(z+x) + z/(x+y)=1 tính M=2019+ x^2/(y+z) + y^2/(z+x) + z^2/(x+y)

Answer 1

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

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

\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)-\left(x+y+z\right)=0\)

\(M=2019\)