Solve: \( \frac{3 x-7}{x^{2}-5 x+6}-\frac{2 x+8}{x^{2}-9}-\frac{x+2}{x^{2}+x-6}=0 \)

To solve the equation \[ \frac{3 x-7}{x^{2}-5 x+6}-\frac{2 x+8}{x^{2}-9}-\frac{x+2}{x^{2}+x-6}=0, \] we start by factoring the denominators. 1. The first denominator: \[ x^2 - 5x + 6 = (x-2)(x-3). \] 2. The second denominator: \[ x^2 - 9 = (x-3)(x+3). \] 3. The third denominator: \[ x^2 + x - 6 = (x-2)(x+3). \] Now we rewrite the equation with these factorizations: \[ \frac{3x-7}{(x-2)(x-3)} - \frac{2x+8}{(x-3)(x+3)} - \frac{x+2}{(x-2)(x+3)} = 0. \] The common denominator will be \((x-2)(x-3)(x+3)\). We rewrite each fraction with the common denominator: 1. First term: \[ \frac{(3x-7)(x+3)}{(x-2)(x-3)(x+3)}. \] 2. Second term: \[ \frac{-(2x+8)(x-2)}{(x-3)(x+3)(x-2)}. \] 3. Third term: \[ \frac{-(x+2)(x-3)}{(x-2)(x+3)(x-3)}. \] Combine them: \[ \frac{(3x-7)(x+3) - (2x+8)(x-2) - (x+2)(x-3)}{(x-2)(x-3)(x+3)} = 0. \] The numerator must equal zero, so we simplify: Expanding each term we have: 1. \( (3x-7)(x+3) = 3x^2 + 9x - 7x - 21 = 3x^2 + 2x - 21 \). 2. \( -(2x+8)(x-2) = -((2x^2 - 4x + 8x - 16)) = -2x^2 + 4x - 8 \) which simplifies to \( -2x^2 + 4x - 8 \). 3. \( -(x+2)(x-3) = -((x^2 - 3x + 2x - 6)) = -x^2 + x - 6 \) which simplifies to \( -x^2 + x - 6 \). Combining all terms: \[ 3x^2 + 2x - 21 - 2x^2 + 4x - 8 - x^2 + x - 6. \] This simplifies to: \[ (3x^2 - 2x^2 - x^2) + (2x + 4x + x) - (21 + 8 + 6) = 0. \] Combining these gives: \[ 0x^2 + 7x - 35 = 0. \] This means: \[ 7x - 35 = 0 \] Solving for \(x\): \[ 7x = 35 \implies x = 5. \] Now, we need to check if this value is valid by ensuring it does not make any denominator zero. The values that make any denominator zero are: 1. \(x - 2 = 0 \rightarrow x = 2\), 2. \(x - 3 = 0 \rightarrow x = 3\), 3. \(x + 3 = 0 \rightarrow x = -3\). As \(x = 5\) does not equal any of these values, it is indeed a valid solution. Thus, the final solution is: \[ \boxed{5}. \]