Question

Cho \(x,y,z\) là ba số thỏa mãn \(x.y.z=1;x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) . Tính giá trị của biểu thức : \(P=\left(x^{19}-1\right)\left(y^{5-1}\right)\left(z^{2016}-1\right)\)

Answer 1

Ta có: \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

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

\(\Leftrightarrow x+y+z=yz+xz+yz\)

Ta có: \(\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

\(=\left(xy-x-y+1\right)\left(z-1\right)\)

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

\(=\left(xyz-1\right)+\left(x+y+z\right)-\left(xy+yz+xz\right)\)

\(=0\)

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

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

Vậy: \(P=\left(x^{19}-1\right)\left(y^5-1\right)\left(z^{2016}-1\right)\)

\(=\left(1^{19}-1\right)\left(1^5-1\right)\left(1^{2016}-1\right)\)

\(=0\)