Okay, let's see. I need to find the sum of the roots a, b, and c of the equation given. The equation is (1/x) + 5x² = 6x - 24. Hmm, first, I remember that for a quadratic equation of the form ax² + bx + c = 0, the sum of the roots is -b/a. But wait, this equation isn't in the standard quadratic form. It has a 1/x term, which is a bit confusing. Maybe I need to rearrange it to make it a standard quadratic equation.

Let me try multiplying both sides by x to eliminate the denominator. If I do that, the left side becomes 1 + 5x³, and the right side becomes 6x² - 24x. So the equation becomes 1 + 5x³ = 6x² - 24x. Then, I can bring all terms to one side to get 5x³ - 6x² + 24x + 1 = 0. Wait, is that right? Let me check again.

Original equation: (1/x) + 5x² = 6x - 24. Multiply both sides by x: 1 + 5x³ = 6x² - 24x. Then subtract 6x² and add 24x to both sides: 5x³ - 6x² + 24x + 1 = 0. Yes, that seems correct. So the cubic equation is 5x³ - 6x² + 24x + 1 = 0.

But the problem states that a, b, c are the roots of this equation. However, the original equation was (1/x) + 5x² = 6x - 24, which is not a polynomial equation. When we multiplied by x, we introduced a term with x³, but does that affect the roots? Wait, when we multiplied by x, we have to be careful because x=0 is a possible solution, but in the original equation, x=0 would make 1/x undefined, so x=0 is not a valid solution. Therefore, the equation after multiplying by x is equivalent to the original equation except for x=0, which isn't a root. So the roots of the original equation are exactly the roots of the cubic equation, excluding x=0. But in the cubic equation, 5x³ -6x² +24x +1=0, the roots are a, b, c (excluding x=0 if it were a root, but since x=0 isn't a root here, because substituting x=0 gives 0 -0 +0 +1=1≠0, so the cubic equation's roots are all the roots of the original equation. Therefore, a, b, c are the three roots of the cubic equation.

But the problem is asking for a + b + c. For a cubic equation of the form px³ + qx² + rx + s = 0, the sum of the roots is -q/p. In our case, the cubic equation is 5x³ -6x² +24x +1=0. So p=5, q=-6. Therefore, the sum of the roots is -q/p = -(-6)/5 = 6/5. So a + b + c = 6/5. Wait, but let me confirm this.

Alternatively, maybe I made a mistake in the rearrangement. Let's check again. Original equation: (1/x) +5x² =6x -24. Multiply both sides by x: 1 +5x³ =6x² -24x. Then bring all terms to left: 5x³ -6x² +24x +1=0. Yes, that's correct. So the cubic equation is indeed 5x³ -6x² +24x +1=0. Therefore, the sum of the roots is -(-6)/5 = 6/5. So the answer is 6/5. But wait, let me make sure that all roots of the original equation are accounted for. Since the original equation is (1/x) +5x² =6x -24, which is equivalent to 1 +5x³ =6x² -24x, so 5x³ -6x² +24x +1=0. Since we multiplied by x, which is not zero, all roots of the original equation are roots of the cubic equation, and vice versa. Therefore, the sum is indeed 6/5. Hmm, seems straightforward. But maybe I should check if the cubic equation has three real roots or something? Wait, but the problem just says a, b, c are the roots, so maybe it's assuming they are real? Or maybe they are complex? But the sum would still be 6/5 regardless. So even if some roots are complex, the sum is still given by Vieta's formula. So regardless of the nature of the roots, the sum is 6/5. Therefore, the answer is 6/5. But let me check once more.

Alternatively, maybe the original equation can be manipulated differently. Let's see. Original equation: (1/x) +5x² =6x -24. Let me rearrange terms: 5x² -6x +24 +1/x =0. Not sure if that helps. Alternatively, maybe express everything in terms of x. Wait, but multiplying by x was the right approach to get rid of the denominator. So I think that was correct. Therefore, the cubic equation is accurate, and the sum of roots is 6/5. So the answer is 6/5. Therefore, the final answer is \boxed{\dfrac{6}{5}}.

But wait, just to be thorough, let me check if there's another way to approach this problem. Suppose I don't convert it to a cubic equation. The original equation is (1/x) +5x² =6x -24. If I let y = x + something, but that might complicate things. Alternatively, maybe there's a substitution. Let me see. Suppose I let t = x - k, trying to eliminate a term. But that might not be necessary here. Alternatively, maybe express the equation as 5x² -6x +24 +1/x =0, but that doesn't seem helpful. So the initial approach of multiplying by x is the correct one, leading to the cubic equation, and then applying Vieta's formula. Therefore, I think the answer is indeed 6/5.
Therefore, the final answer is $\boxed{\frac{6}{5}}$.
</think>