Oh variables

$x+y=2z \Leftrightarrow y=2z-x$

$\Rightarrow \dfrac{x}{x-z}+\dfrac{z}{y-z} \Leftrightarrow \dfrac{x}{x-z}+\dfrac{z}{2z-x-z}$

$\Leftrightarrow \dfrac{x}{x-z}+\dfrac{z}{z-x}$

$\Leftrightarrow \dfrac{x}{x-z}-\dfrac{z}{x-z}=\dfrac{x-z}{x-z}=\boxed{1}$

$\square$

$\dfrac{x}{x-z}+\dfrac{z}{y-z} = \dfrac{x(y-z)+z(x-z)}{(x-y)(x-z)} = \dfrac{xy-xz+xz-z^2}{xy-z(x+y)+z^2}$

We just need to simplify and use $x+y = 2z$ , then we will get $\frac{xy-z^2}{xy-z(2z)+z^2} = \frac{xy-z^2}{xy-z^2} = 1$

we have $x+y=2z$ .Let's try to obtain some equations from $x+y=2z$ . $x-z=z-y......1 \\ x=2z-y......2$ simplifying $\dfrac{x}{x-z}+\dfrac{z}{y-z}$ ,we get $\dfrac{xy-z^{2}}{(x-z)(y-z)}=\dfrac{xy-z^{2}}{-(y-z)^{2}}....because (x-z)=(z-y) \\ =\dfrac{(2z-y)y-z^{2}}{-(y-z)^{2}}...from2 \\ =\dfrac{2zy-y^{2}-z^{2}}{-(y-z)^{2}} \\ =\dfrac{-(y-z)^{2}}{-(y-z)^{2}}=\boxed{1}$

We have, $x+y=2z$ So, $y-z = z-x$ Now, $\dfrac{x}{x-z} + \dfrac{z}{y-z}$ $=\dfrac{-x}{z-x} + \dfrac{z}{y-z}$ $=\dfrac{-x}{z-x} + \dfrac{z}{z-x}$ $= \dfrac{-x+z}{z-x}$ $= \boxed {1}$

given that x+y=2z.... x-z=z-y=k.... (x/k)-(y/k)=(x-y)/k=k/k=1

x+y= 2z,then y = 2z-x, so (x/x-z) + (z/y-x) = (x/x-z) + (z/2z-x-z) = (x/x-z) + (z/z-x) = (x/x-z) - (z/x-x) = x-z/x-z =1

x+y=2z =>y=2z-x y-z becomes z-x or -(x-z) So z/(y-z) becomes -z/(x-z) Adding x/(x-z) and -z/(x-z) gives 1