Question

Cho xyz=1.Tính giá trị biểu thức P=\(\frac{1}{1+x+xy}\)+\(\frac{1}{1+y+yz}\)+\(\frac{1}{1+z+xz}\)

Answer 1

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

\(P=\frac{1}{1+x+xy}+\frac{x}{x\left(1+y+yz\right)}+\frac{xy}{xy\left(1+z+xz\right)}\)

\(P=\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+xyz+x^2yz}\)

\(P=\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+xyz+xyz.x}\)

\(P=\frac{1}{1+x+xy}+\frac{x}{x+xy+1}+\frac{xy}{xy+1+x}\left(xyz=1\right)\)

\(P=\frac{1+x+xy}{1+x+xy}=1\)

Vậy P=1