Question

Tính tổng ba phân thức sau:                                     x/(xy+x+1)+y/(yz+y+1)+z/(zx+z+1)​                  Biết xyz=1

Answer 1

moi hok lop 6

Answer 2

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

\(=\frac{x}{x.\left(yz+y+1\right)}+\frac{y}{yz+z+1}+\frac{yz}{yz+y+1}=\frac{1}{yz+y+1}+\frac{y}{yz+z+1}+\frac{yz}{yz+y+1}\)

\(=\frac{1+y+yz}{yz+y+1}=1\)