If `5x^2-2kx+1 < 0` has exactly one integral solution which is `1` then sum of all positive integral values of k.

Answer

To solve the inequality \(5x^2 - 2kx + 1 < 0\) which has exactly one integral solution at \(x = 1\), we can follow these steps:

### Step 1: Substitute the integral solution into the equation

Since \(x = 1\) is a solution, we can substitute \(x = 1\) into the quadratic equation:

\[ 5(1)^2 - 2k(1) + 1 = 0 \]

This simplifies to:

\[ 5 - 2k + 1 = 0 \]

Combining like terms gives:

\[ 6 - 2k = 0 \]

### Step 2: Solve for \(k\)

Rearranging the equation gives:

\[ 2k = 6 \implies k = 3 \]

### Step 3: Determine conditions for the inequality to hold

For the quadratic \(5x^2 - 2kx + 1 < 0\) to have exactly one integral solution, the discriminant must be zero (indicating a double root). The discriminant \(D\) of the quadratic \(ax^2 + bx + c\) is given by:

\[ D = b^2 - 4ac \]

For our quadratic, \(a = 5\), \(b = -2k\), and \(c = 1\):

\[ D = (-2k)^2 - 4(5)(1) = 4k^2 - 20 \]

Setting the discriminant equal to zero for the double root condition:

\[ 4k^2 - 20 = 0 \]

### Step 4: Solve for \(k\) again

Rearranging gives:

\[ 4k^2 = 20 \implies k^2 = 5 \implies k = \sqrt{5} \text{ or } k = -\sqrt{5} \]

### Step 5: Check the conditions for integral solutions

Since we need \(k\) to be a positive integer, we need to find values of \(k\) such that \(5x^2 - 2kx + 1 < 0\) has exactly one integral solution at \(x = 1\).

### Step 6: Analyze the range of \(k\)

From the conditions derived, we have:

1. \(k^2 > 5\) (from the discriminant being positive)
2. \(k^2 < 30\) (from the condition on the roots)

This means:

\[ \sqrt{5} < k < \sqrt{30} \]

Calculating the approximate values:

- \(\sqrt{5} \approx 2.236\)
- \(\sqrt{30} \approx 5.477\)

Thus, the possible positive integral values of \(k\) are \(3, 4, 5\).

### Step 7: Verify which values satisfy the original inequality

We need to check which of these values satisfy the condition that \(5x^2 - 2kx + 1 < 0\) has exactly one integral solution at \(x = 1\).

- For \(k = 3\):
- The quadratic becomes \(5x^2 - 6x + 1\). The discriminant is \(0\), so it has a double root at \(x = 1.2\) (not an integer).

- For \(k = 4\):
- The quadratic becomes \(5x^2 - 8x + 1\). The discriminant is \(0\), so it has a double root at \(x = 0.8\) (not an integer).

- For \(k = 5\):
- The quadratic becomes \(5x^2 - 10x + 1\). The discriminant is \(0\), so it has a double root at \(x = 1\) (which is an integer).

### Conclusion

The only valid positive integral value of \(k\) that satisfies all conditions is \(k = 5\).

### Final Answer

The sum of all positive integral values of \(k\) is:

\[ \text{Sum} = 5 \]