Find the real number $t$, such that the following system of equations has a unique real solution $(x, y, z, v)$:

\[ \left\{\begin{array}{cc}x+y+z+v=0\\ (xy + yz +zv)+t(xz+xv+yv)=0\end{array}\right. \]
1. We start with the given system of equations:
   \[
   \left\{
   \begin{array}{l}
   x + y + z + v = 0 \\
   (xy + yz + zv) + t(xz + xv + yv) = 0
   \end{array}
   \right.
   \]

2. Substitute \( v = -x - y - z \) into the second equation:
   \[
   (xy + yz + z(-x - y - z)) + t(xz + x(-x - y - z) + y(-x - y - z)) = 0
   \]
   Simplify the terms:
   \[
   xy + yz - xz - yz - z^2 + t(xz - x^2 - xy - xz - xy - y^2 - yz) = 0
   \]
   \[
   xy - z^2 + t(-x^2 - 2xy - y^2 - yz) = 0
   \]
   \[
   xy - z^2 - t(x^2 + 2xy + y^2 + yz) = 0
   \]

3. Rearrange the equation:
   \[
   z^2 + t(x^2 + 2xy + y^2 + yz) = xy
   \]
   \[
   z^2 + t(x^2 + 2xy + y^2 + yz) - xy = 0
   \]

4. To ensure the quadratic equation in \( z \) has real roots, the discriminant must be non-negative:
   \[
   \Delta = (x + ty)^2 - 4t(x^2 + 2xy + y^2 - xy)
   \]
   Simplify the discriminant:
   \[
   \Delta = (x + ty)^2 - 4t(x^2 + 2xy + y^2 - xy)
   \]
   \[
   \Delta = x^2 + 2txy + t^2y^2 - 4t(x^2 + 2xy + y^2 - xy)
   \]
   \[
   \Delta = x^2 + 2txy + t^2y^2 - 4tx^2 - 8txy - 4ty^2 + 4txy
   \]
   \[
   \Delta = x^2 + 2txy + t^2y^2 - 4tx^2 - 4txy - 4ty^2
   \]
   \[
   \Delta = x^2(1 - 4t) + 2xy(t - 2t) + y^2(t^2 - 4t)
   \]
   \[
   \Delta = x^2(1 - 4t) - 2txy + y^2(t^2 - 4t)
   \]

5. For the quadratic to have a unique solution, the discriminant must be zero:
   \[
   x^2(1 - 4t) - 2txy + y^2(t^2 - 4t) = 0
   \]

6. Analyze the conditions for \( t \):
   - If \( t < \frac{1}{4} \), the inequality \( x^2(1 - 4t) - 2txy + y^2(t^2 - 4t) \geq 0 \) has infinitely many solutions.
   - If \( t = \frac{1}{4} \), the inequality becomes \( (8x - 3y)y \geq 0 \), which also has infinitely many solutions.
   - Therefore, we need \( t > \frac{1}{4} \).

7. Further analyze the discriminant condition:
   \[
   4y^2(2 - 3t)^2 - 4y^2(1 - 4t)(t^2 - 4t) \geq 0
   \]
   \[
   16y^2(t + 1)(t^2 - 3t + 1) \geq 0
   \]

8. For the inequality \( (t + 1)(t^2 - 3t + 1) \geq 0 \) to hold, we need:
   \[
   t \in \left( \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2} \right)
   \]

9. Therefore, the value of \( t \) that ensures a unique real solution is:
   \[
   t \in \left( \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2} \right)
   \]

The final answer is \( \boxed{ t \in \left( \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2} \right) } \)