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To solve this problem, we need to follow two steps:

### Given:
The function is \( y = \frac{\sqrt{x}}{x + 5} \).

We are tasked with determining:
1. The equation of the tangent line at the point \( \left(9, \frac{3}{14} \right) \).
2. The equation of the normal line at the same point.

### Step 1: Tangent Line
The equation of the tangent line at a point \( (x_1, y_1) \) is given by:
\[
y - y_1 = m_t(x - x_1)
\]
Where \( m_t \) is the slope of the tangent line, which is found by differentiating the given function.

#### Differentiate \( y = \frac{\sqrt{x}}{x + 5} \)

We can apply the quotient rule for differentiation:

\[
\frac{d}{dx} \left( \frac{\sqrt{x}}{x + 5} \right) = \frac{(x + 5) \cdot \frac{d}{dx}(\sqrt{x}) - \sqrt{x} \cdot \frac{d}{dx}(x+5)}{(x + 5)^2}
\]

- The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).
- The derivative of \( x + 5 \) is 1.

Thus, the derivative (slope of the tangent line) is:
\[
\frac{d}{dx} \left( \frac{\sqrt{x}}{x + 5} \right) = \frac{(x + 5) \cdot \frac{1}{2\sqrt{x}} - \sqrt{x}}{(x + 5)^2}
\]

#### Evaluate the derivative at \( x = 9 \)

Substitute \( x = 9 \) into the derivative to find the slope of the tangent line \( m_t \).

\[
m_t = \frac{(9 + 5) \cdot \frac{1}{2\sqrt{9}} - \sqrt{9}}{(9 + 5)^2}
\]
\[
m_t = \frac{14 \cdot \frac{1}{6} - 3}{14^2}
\]
\[
m_t = \frac{\frac{14}{6} - 3}{196} = \frac{\frac{7}{3} - 3}{196} = \frac{\frac{7}{3} - \frac{9}{3}}{196} = \frac{-\frac{2}{3}}{196} = -\frac{2}{588} = -\frac{1}{294}
\]

Thus, the slope of the tangent line is \( m_t = -\frac{1}{294} \).

Now, we use the point-slope form of the tangent line equation with the point \( \left(9, \frac{3}{14}\right) \):

\[
y - \frac{3}{14} = -\frac{1}{294}(x - 9)
\]
Simplify to get the equation of the tangent line.

### Step 2: Normal Line
The slope of the normal line is the negative reciprocal of the tangent line slope. Since \( m_t = -\frac{1}{294} \), the slope of the normal line \( m_n \) is:

\[
m_n = 294
\]

Now use the point-slope form for the normal line:

\[
y - \frac{3}{14} = 294(x - 9)
\]
Simplify to get the equation of the normal line.

Would you like me to continue simplifying these equations?

Determine the equation of the tangent line and the normal line at the point (9, 3/14) for the function y = √x / (x + 5).

This problem demonstrates how to determine the equations of the tangent and normal lines for the function y = √x / (x + 5) at the point (9, 3/14). Using differentiation, including the quotient rule, this solution provides a step-by-step explanation suitable for advanced high school or college calculus students.

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