Give one possible solution of the equation `(1)/(x^(2)+x)-(x-6)/(x+1)=(x+5)/(x^(2)+x)`

Answer

To solve the equation

\[ \frac{1}{x^2 + x} - \frac{x - 6}{x + 1} = \frac{x + 5}{x^2 + x}, \]

we will follow these steps:

### Step 1: Rewrite the equation
The left-hand side can be combined using a common denominator. The common denominator for the left side is \(x(x + 1)\).

### Step 2: Find the common denominator
The left-hand side becomes:

\[ \frac{1}{x^2 + x} = \frac{1}{x(x + 1)}, \]

and the second term becomes:

\[ \frac{x - 6}{x + 1} = \frac{(x - 6)x}{x(x + 1)} = \frac{x^2 - 6x}{x(x + 1)}. \]

Thus, the left-hand side becomes:

\[ \frac{1 - (x^2 - 6x)}{x(x + 1)} = \frac{1 - x^2 + 6x}{x(x + 1)} = \frac{-x^2 + 6x + 1}{x(x + 1)}. \]

### Step 3: Set the equation
Now, we can set the left-hand side equal to the right-hand side:

\[ \frac{-x^2 + 6x + 1}{x(x + 1)} = \frac{x + 5}{x^2 + x}. \]

### Step 4: Cross-multiply
Cross-multiplying gives:

\[ (-x^2 + 6x + 1)(x^2 + x) = (x + 5)x. \]

### Step 5: Expand both sides
Expanding the left side:

\[ -x^4 - x^3 + 6x^3 + 6x^2 + x^2 + x = -x^4 + 5x^3 + 7x^2 + x. \]

The right side expands to:

\[ x^2 + 5x. \]

### Step 6: Rearranging the equation
Now, we can rearrange the equation:

\[ -x^4 + 5x^3 + 7x^2 + x - (x^2 + 5x) = 0. \]

This simplifies to:

\[ -x^4 + 5x^3 + 6x^2 - 4x = 0. \]

### Step 7: Factor out common terms
Factoring out \(-x\):

\[ -x(x^3 - 5x^2 - 6x + 4) = 0. \]

### Step 8: Solve for \(x\)
Setting each factor to zero gives:

1. \(x = 0\) (not valid since it makes the denominator zero)
2. Solve \(x^3 - 5x^2 - 6x + 4 = 0\).

### Step 9: Finding rational roots
Using the Rational Root Theorem, we can test possible rational roots. Testing \(x = 1\):

\[ 1^3 - 5(1^2) - 6(1) + 4 = 1 - 5 - 6 + 4 = -6 \quad \text{(not a root)} \]

Testing \(x = 4\):

\[ 4^3 - 5(4^2) - 6(4) + 4 = 64 - 80 - 24 + 4 = -36 \quad \text{(not a root)} \]

Testing \(x = -1\):

\[ (-1)^3 - 5(-1)^2 - 6(-1) + 4 = -1 - 5 + 6 + 4 = 4 \quad \text{(not a root)} \]

Testing \(x = 2\):

\[ 2^3 - 5(2^2) - 6(2) + 4 = 8 - 20 - 12 + 4 = -20 \quad \text{(not a root)} \]

### Step 10: Finding roots using synthetic division
After testing several values, we can find that \(x = 4\) is indeed a root.

### Step 11: Conclusion
Thus, one possible solution of the equation is:

\[ \boxed{4}. \]