Question

cho x,y,z thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)CM: \(\frac{1}{x^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{x^{2019}+y^{2019}+z^{2019}}\)

Answer 1

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)=> \(\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)

=> (x+y+z)(xy+yz+zx) = xyz

=> \(x^2y+xy^2+y^2z+yz^2+zx^2+z^2x+2xyz=0\)

=> (x+y)(y+z)(z+x) = 0

=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

TH1: x = -y

=> \(\frac{1}{x^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{\left(-y\right)^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{z^{2019}}\)

=> \(\frac{1}{x^{2019}+y^{2019}+z^{2019}}=\frac{1}{\left(-y\right)^{2019}+y^{2019}+z^{2019}}=\frac{1}{z^{2019}}\)

=> ĐPCM

Tương tự với TH2 và TH3