olve \( \sqrt{x+5}+1=\frac{6}{\sqrt{x+5}} \)

To solve the equation \( \sqrt{x+5} + 1 = \frac{6}{\sqrt{x+5}} \), we'll start by making a substitution. Let \( y = \sqrt{x+5} \). Then the equation becomes: \[ y + 1 = \frac{6}{y} \] Next, we can eliminate the fraction by multiplying both sides by \( y \) (assuming \( y \neq 0 \)): \[ y^2 + y = 6 \] Rearranging this gives us a quadratic equation: \[ y^2 + y - 6 = 0 \] Now, we can factor this equation: \[ (y - 2)(y + 3) = 0 \] Setting each factor to zero gives us: \[ y - 2 = 0 \quad \text{or} \quad y + 3 = 0 \] So, we have: \[ y = 2 \quad \text{or} \quad y = -3 \] Since \( y = \sqrt{x+5} \) must be non-negative, we discard \( y = -3 \). Thus, we have: \[ y = 2 \] Now, substituting back to find \( x \): \[ \sqrt{x+5} = 2 \] Squaring both sides: \[ x + 5 = 4 \] Solving for \( x \): \[ x = 4 - 5 = -1 \] Finally, we should check our solution by substituting \( x = -1 \) back into the original equation: \[ \sqrt{-1 + 5} + 1 = \frac{6}{\sqrt{-1 + 5}} \] Calculating this gives: \[ \sqrt{4} + 1 = \frac{6}{\sqrt{4}} \] This simplifies to: \[ 2 + 1 = \frac{6}{2} \] Which results in: \[ 3 = 3 \] This confirms our solution works. Therefore, the solution to the equation \( \sqrt{x+5}+1=\frac{6}{\sqrt{x+5}} \) is: \[ \boxed{-1} \]