Question

Biết \(xyz=1\), hãy tính tổng: 

\(A=\dfrac{5}{x+xy+1}+\dfrac{5}{y+yz+1}+\dfrac{5}{z+zx+1}\)

Answer 1

Ta có: \(A=\frac{5}{x+xy+1}+\frac{5}{y+yz+1}+\frac{5}{z+zx+1}\)

\(=5\left(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\right)\)

\(=5\left(\frac{z}{xz+xyz+z}+\frac{xz}{xyz+xyz^2+xz}+\frac{1}{z+zx+1}\right)=5\left(\frac{z}{xz+z+1}+\frac{xz}{1+z+xz}+\frac{1}{z+xz+1}\right)=5\cdot\frac{z+xz+1}{z+xz+1}=5\)