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There are no real solutions:

From the first equation we see that $x=a$ where we introduced $\quad a = \exp(-1+x+y+z)$. Now we get from the second equation $y=a \exp(-x/2)$ and in turn from the first one we have $y=x \exp(-x/2) $. From the third equation we get $z=a\exp(-y/5)$. Next we can obtain an equation for x by eliminating $y$ and $z$ i.e. $$x=e^{-1+x+x e^{-\frac{x}{2}} + x e^{-\frac{x}{5} e^{-\frac{x}{2}}}}$$ We need to restrict appropriately the complex domain since knowing properties of exponential functions we expect there are infinitely many solutions. Now we obtain two exact solutions for x nearest zero in the complex plane:

Root is a symbolic representation of exact solutions, see e.g. How do I work with Root objects?. For example taking restriction Abs[z]<5 after a few minutes I got 133 complex solutions for x. One can show that if x + I y is a solution then x - I y is a solution too (for x and y reals)

Let's demonstrate the structure of the x-solution space in a small part of the complex plane. Defining

we can catch a glimpse of multitude of solutions

ContourPlot[{Re[f[x, y]] == 0, Im[f[x, y]] == 0},
            {x, -4.1, -2.9}, {y, 0.4, 1.6}, PlotPoints -> 50,
             MaxRecursion -> 2, ImageSize -> Large]

Solutions are where orange ang blue cureves intersect.

{y -> 0.177717 - 0.466416 I, y -> 0.177717 + 0.466416 I}

There are infinitely many solutions. Here we demonstrate only a small part of the x - solution space in the complex plane.

Let's start from the begining. There are no real solutions:

From the first equation we see that $x=a$ where we introduced $\quad a = \exp(-1+x+y+z)$. Now we get from the second equation $y=a \exp(-x/2)$ and in turn from the first one we have $y=x \exp(-x/2) $. From the third equation we get $z=a\exp(-y/5)$. Next we can obtain an equation for $x$ by eliminating $y$ and $z$ i.e. it yields $$x=e^{-1+x+x e^{-\frac{x}{2}} + x e^{-\frac{x}{5} e^{-\frac{x}{2}}}}$$ We need to restrict appropriately the complex domain since knowing properties of exponential functions we expect there are infinitely many solutions. Now we obtain two exact solutions for x nearest zero in the complex plane:

Root is a symbolic representation of exact solutions, see e.g. How do I work with Root objects?. For example taking restriction Abs[x] < 370/100 after a few minutes I got 102 complex solutions for x. One can show that if x + I y is a solution then x - I y is a solution too (for x and y reals)

To demonstrate the structure of the x-solution space in a small part of the complex plane we define:

and we can catch a glimpse of multitude of solutions

sol = {ToRules@
Reduce[x == 
   Exp[-1 + x (1 + Exp[-x/2] + Exp[-(x/5) Exp[-(x/2)]])] && 
  Abs[x] < 370/100, x]};
Length[sol]

102

ContourPlot[{Re[f[x, y]] == 0, Im[f[x, y]] == 0},
            {x, -3.7, -2.85}, {y, 0.65, 1.5}, 
  PlotPoints -> 50, ImageSize -> Large,  
  ContourStyle -> {Orange, Darker@Cyan},  
  Epilog -> {Red, PointSize[0.015], Point[ReIm[x /. sol]]}]

Solutions are denoted by red points where curves intersect. For a technical reason not all the intersection point are drawn.

{y -> 0.177717 - 0.466416 I, y -> 0.177717 + 0.466416 I}

If we are looking for real solutions, there are none.

Without domain specification (it means we are looking for complex solutions) Reduce has been working for several minutes without yielding results. This is not surprising. Instead we take a pedestrian approach, although it might be automized a bit more.

From the first equation we see that $x=a$ where we introduced $\quad a = \exp(-1+x+y+z)$. Now we have from the second equation we get $y=a \exp(-x/2) $ and from the third equation we get $z=a\exp(-y/5)$. Next we get an equation for x by eliminating $y$ and $z$ i.e. $x=e^{-1+x+x e^{-\frac{x}{2}} + x e^{-\frac{x}{5} e^{-\frac{x}{2}}}}$. We need to restrict appropriately the complex domain since knowing properties of exponential functions we expect there are infinitely many solutions.

The above we've got symbolic representation of exact solutions. For example taking restriction Abs[z]<5 after roughly 3 minutes I got 133 complex solutions for x. The results for y and z might be expressed also in the symbolic form, nontheless for brevity we demonstrate them only in the numerical form

There are no real solutions:

Without domain specification (it means we are looking for complex solutions) Reduce has to be supplemented by an appropriate domain specification, otherwise it won't find any solutions.

To demonstrate what is going on here let's take a pedestrian approach, after all it might be automized a bit.

From the first equation we see that $x=a$ where we introduced $\quad a = \exp(-1+x+y+z)$. Now we get from the second equation $y=a \exp(-x/2)$ and in turn from the first one we have $y=x \exp(-x/2) $. From the third equation we get $z=a\exp(-y/5)$. Next we can obtain an equation for x by eliminating $y$ and $z$ i.e. $$x=e^{-1+x+x e^{-\frac{x}{2}} + x e^{-\frac{x}{5} e^{-\frac{x}{2}}}}$$ We need to restrict appropriately the complex domain since knowing properties of exponential functions we expect there are infinitely many solutions. Now we obtain two exact solutions for x nearest zero in the complex plane:

Root is a symbolic representation of exact solutions, see e.g. How do I work with Root objects?. For example taking restriction Abs[z]<5 after a few minutes I got 133 complex solutions for x. One can show that if x + I y is a solution then x - I y is a solution too (for x and y reals)

Let's demonstrate the structure of the x-solution space in a small part of the complex plane. Defining

f[x_, y_] := -E^(-1 + (1 + E^(1/2 (-x - I y)) + 
   E^(-(1/5) E^(1/2 (-x - I y)) (x + I y))) (x + I y)) + x + I y

we can catch a glimpse of multitude of solutions

ContourPlot[{Re[f[x, y]] == 0, Im[f[x, y]] == 0},
            {x, -4.1, -2.9}, {y, 0.4, 1.6}, PlotPoints -> 50,
            MaxRecursion -> 2, ImageSize -> Large]

Solutions are where orange ang blue cureves intersect.

The results for y and z might be expressed also in the symbolic form, nontheless for brevity we demonstrate them only related to given two solutions for x in the numerical form

Without domain specification Reduce has been workin for several minutes without yielding results. This is not quite surprising. Instead we take a pedestrian approach, although it might be automized a bit more.

From the first equation we see that $x=\exp(a)$ where we introduced $\quad a = -1+x+y+z$. Now we have from the second equation we get $y=\exp(a) \exp(-x/2)$ and from the third equation we get $z=\exp(a) \exp(-y/5)$. Next we get an equation for x by eliminating $y$ and $z$. We need to restrict the domain

Reduce[ x == Exp[-1 + x (1 + Exp[-x/2] + Exp[-(x/5) Exp[-(x/2)]])] && 
        Abs[x] < 2, x]

Knowing properties of the exponential function we be beliieve there are infinitely many solutions. For example taking restriction Abs[z]<5 after roughly 3 minutes I got 133 solutions. Getting results for y and z is an exercise for you since you haven't demonstrated much of your Mathematica skills.

Without domain specification (it means we are looking for complex solutions) Reduce has been working for several minutes without yielding results. This is not surprising. Instead we take a pedestrian approach, although it might be automized a bit more.

From the first equation we see that $x=a$ where we introduced $\quad a = \exp(-1+x+y+z)$. Now we have from the second equation we get $y=a \exp(-x/2) $ and from the third equation we get $z=a\exp(-y/5)$. Next we get an equation for x by eliminating $y$ and $z$ i.e. $x=e^{-1+x+x e^{-\frac{x}{2}} + x e^{-\frac{x}{5} e^{-\frac{x}{2}}}}$. We need to restrict appropriately the complex domain since knowing properties of exponential functions we expect there are infinitely many solutions.

Reduce[ x == Exp[-1 + x (1 + Exp[-x/2] + Exp[-(x/5) Exp[-(x/2)]])] && 
        Abs[x] < 2, x]
sx = {ToRules @ % } // N;

The above we've got symbolic representation of exact solutions. For example taking restriction Abs[z]<5 after roughly 3 minutes I got 133 complex solutions for x. The results for y and z might be expressed also in the symbolic form, nontheless for brevity we demonstrate them only in the numerical form

sy = y -> x Exp[-x/2] /. sx

{y -> 0.177717 - 0.466416 I, y -> 0.177717 + 0.466416 I}

z -> x  Exp[-y/5] /. Transpose[{sx // Flatten, sy}]

{z -> 0.0994856 - 0.485233 I, z -> 0.0994856 + 0.485233 I}

Reduce couldn't find solutions without approprate restriction of domains for x, y, z separately. This is crucial point to get any solutions.