x(x-2)/x+3<0

To solve the inequality \(\frac{x(x-2)}{x+3} < 0\), we will follow these steps:

1. Find critical points: The critical points can be found by determining where the numerator and denominator are equal to zero.

   • Numerator: \(x(x-2) = 0\)

     • Solutions are \(x = 0\) and \(x = 2\).

   • Denominator: \(x + 3 = 0\)

     • Solution is \(x = -3\).

   So the critical points are \(x = -3\), \(x = 0\), and \(x = 2\).

2. Create intervals: The critical points divide the number line into the following intervals:

   • \((- \infty, -3)\)
   • \((-3, 0)\)
   • \((0, 2)\)
   • \((2, \infty)\)

3. Test each interval: We will choose a test point from each interval to determine if the expression is negative in that interval.

   • Interval \((- \infty, -3)\): Choose \(x = -4\) \[ \frac{-4(-4-2)}{-4+3} = \frac{-4(-6)}{-1} = \frac{24}{-1} = -24 < 0 \]

   • Interval \((-3, 0)\): Choose \(x = -1\) \[ \frac{-1(-1-2)}{-1+3} = \frac{-1(-3)}{2} = \frac{3}{2} > 0 \]

   • Interval \((0, 2)\): Choose \(x = 1\) \[ \frac{1(1-2)}{1+3} = \frac{1(-1)}{4} = \frac{-1}{4} < 0 \]

   • Interval \((2, \infty)\): Choose \(x = 3\) \[ \frac{3(3-2)}{3+3} = \frac{3(1)}{6} = \frac{3}{6} = \frac{1}{2} > 0 \]

4. Conclusion from test points: The expression is negative in the intervals \((- \infty, -3)\) and \((0, 2)\).

5. Include/exclude critical points:

   • At \(x = -3\), the denominator is zero, so it is not included.
   • At \(x = 0\) and \(x = 2\), the entire expression equals zero, which does not satisfy the strict inequality.

6. Final solution: The solution to the inequality \(\frac{x(x-2)}{x+3} < 0\) is: \[ \boxed{(-\infty, -3) \cup (0, 2)} \]