Chứng minh rằng nếu ta có `xyz=1` thi ta cũng có `\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}=1` 18/09/2021 Bởi Audrey Chứng minh rằng nếu ta có `xyz=1` thi ta cũng có `\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}=1`
$\quad \dfrac{x}{xy + x + 1} + \dfrac{y}{yz + y + 1} + \dfrac{z}{zx + z+1}$ $=\dfrac{xyz}{xy^2z + xyz + yz} +\dfrac{y}{yz + y + 1} + \dfrac{yz}{xyz + yz + y}$ $=\dfrac{1}{y + 1 + yz} +\dfrac{y}{yz + y + 1} +\dfrac{yz}{1 + yz + y}$ $=\dfrac{yz + y +1}{yz + y + 1}$ $= 1$ Bình luận
$\quad \dfrac{x}{xy + x + 1} + \dfrac{y}{yz + y + 1} + \dfrac{z}{zx + z+1}$
$=\dfrac{xyz}{xy^2z + xyz + yz} +\dfrac{y}{yz + y + 1} + \dfrac{yz}{xyz + yz + y}$
$=\dfrac{1}{y + 1 + yz} +\dfrac{y}{yz + y + 1} +\dfrac{yz}{1 + yz + y}$
$=\dfrac{yz + y +1}{yz + y + 1}$
$= 1$