Question

Cho x, y, z thỏa mãn: xyz=1

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

Tính\(P=\left(x^{19}-1\right)\left(y^5-1\right)\left(z^{1890}-1\right)\)

Answer 1

Lời giải:

Từ điều kiện $xyz=1$ ta có:

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

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

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

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

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

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

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

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

Khi đó:
\(P=(x^{19}-1)(y^5-1)(z^{1890}-1)=(x-1)(x^{18}+x^{17}+...+1)(y-1)(y^4+...+1)(z-1)(z^{1889}+...+1)\)

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