1) \( x^{2}+5 x+2=\frac{24}{x^{2}+5 x} \)

To solve the equation \( x^{2}+5 x+2=\frac{24}{x^{2}+5 x} \), we start by multiplying both sides by \( x^{2} + 5x \) (as long as it’s not zero) to eliminate the fraction: \[ (x^{2}+5x)(x^{2}+5x+2) = 24 \] Next, we let \( y = x^{2} + 5x \). The equation simplifies to: \[ y(y + 2) = 24 \implies y^{2} + 2y - 24 = 0 \] This is a quadratic equation that we can solve using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^{2} - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2} = \frac{-2 \pm 10}{2} \] This gives us: \[ y = \frac{8}{2} = 4 \quad \text{or} \quad y = \frac{-12}{2} = -6 \] Now, substituting \( y = x^{2} + 5x \): 1. For \( y = 4 \): \[ x^{2} + 5x - 4 = 0 \] Using the quadratic formula again: \[ x = \frac{-5 \pm \sqrt{25 + 16}}{2} = \frac{-5 \pm \sqrt{41}}{2} \] 2. For \( y = -6 \): \[ x^{2} + 5x + 6 = 0 \] Solving this one yields: \[ x = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2} \implies x = -2 \quad \text{or} \quad x = -3 \] So, the solutions are: \[ x = \frac{-5 + \sqrt{41}}{2}, \quad x = \frac{-5 - \sqrt{41}}{2}, \quad x = -2, \quad x = -3 \]