Question

cho xyz=1 và x +y+z=\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

tính P = (x\(^{1999}\)-1)(y\(^{2018}\)-1)(z\(^{2019}\)-1)

Answer 1

Lời giải:
Vì $xyz=1$ nên:

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

\(\Leftrightarrow x+y+z-xy-yz-xz=0\)

\(\Leftrightarrow 1+x+y+z-xy-yz-xz-1=0\)

\(\Leftrightarrow xyz+x+y+z-xy-yz-xz-1=0\)

\(\Leftrightarrow xy(z-1)+(x+y-yz-xz)+(z-1)=0\)

\(\Leftrightarrow xy(z-1)-x(z-1)-y(z-1)+(z-1)=0\)

\(\Leftrightarrow (z-1)(xy-x-y+1)=0\)

\(\Leftrightarrow (z-1)(x-1)(y-1)=0\)

Do đó:

\(P=(x^{1999}-1)(y^{2018}-1)(z^{2019}-1)\)

\(=(x-1)(x^{1998}+x^{1997}+...+1)(y-1)(y^{2017}+...+1)(z-1)(z^{2018}+....+1)\)

\(=(x-1)(y-1)(z-1).A=0.A=0\)