8. Specify the smallest integer value of \(a\) for which the system has a unique solution

\[
\left\{\begin{array}{l}
\frac{y}{a-\sqrt{x}-1}=4 \\
y=\frac{\sqrt{x}+5}{\sqrt{x}+1}
\end{array}\right.
\]
# Solution.

Solving the system by substitution, we arrive at an equation with constraints on the unknown quantity ${ }^{x}$.

$$
\left\{\begin{array} { l } 
{ \frac { y } { a - \sqrt { x } - 1 } = 4 } \\
{ y = \frac { \sqrt { x } + 5 } { \sqrt { x } + 1 } }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x \geq 0 } \\
{ \sqrt { x } \neq a - 1 } \\
{ \frac { \sqrt { x } + 5 } { \sqrt { x } + 1 } = 4 ( a - \sqrt { x } - 1 ) }
\end{array} \Rightarrow \left\{\begin{array}{l}
x \geq 0 \\
\sqrt{x} \neq a-1 \\
\sqrt{x}+5=4(a-\sqrt{x}-1)(\sqrt{x}+1)
\end{array}\right.\right.\right.
$$

First, let's check for which values of the parameter the case $\sqrt{x}=a-1$ is possible. $4(a-1)^{2}+(9-4 a)(a-1)+(9-4 a)=0 \Rightarrow 4(a-1)^{2}+9 a-4 a^{2}=0 \Rightarrow a=-4$.

For this value of the parameter, we get a root equal to a negative number, which is impossible.

We solve the quadratic equation $4 x+(9-4 a) \sqrt{x}+(9-4 a)=0$.

$D=(4 a-9)(4 a+7) \Rightarrow \sqrt{x}_{1,2}=\frac{4 a-9 \pm \sqrt{(4 a-9)(4 a+7)}}{8}$.

The only non-negative solution will be under the conditions

$$
\left[\begin{array}{l}
\left\{\begin{array}{l}
(4 a-9)(4 a+7)=0 \\
\frac{4 a-9}{8} \geq 0
\end{array}\right. \\
\left\{\begin{array} { l } 
{ a \in ( - \infty , - 7 / 4 ) \cup ( 9 / 4 , + \infty ) } \\
{ [ \begin{array} { l } 
{ \frac { 9 - 4 a } { 4 } < 0 }
\end{array} } \\
{ \{ \begin{array} { l } 
{ x _ { 1 } x _ { 2 } = \frac { 9 - 4 a } { 4 } = 0 } \\
{ x _ { 1 } + x _ { 2 } = \frac { 4 a - 9 } { 4 } < 0 }
\end{array} }
\end{array} \Rightarrow \left[\begin{array}{l}
a=9 / 4 \\
a \in(9 / 4,+\infty)
\end{array}\right.\right.
\end{array}\right.
$$

Choosing the smallest integer value of the parameter, we get $a=3$.

## Answer: 3.