For which of the following values of real number $t$, the equation $x^4-tx+\dfrac 1t = 0$ has no root on the interval $[1,2]$?

$ 
\textbf{(A)}\ 6
\qquad\textbf{(B)}\ 7
\qquad\textbf{(C)}\ 8
\qquad\textbf{(D)}\ 9
\qquad\textbf{(E)}\ \text{None of the preceding}
$
To determine for which values of \( t \) the equation \( x^4 - tx + \frac{1}{t} = 0 \) has no root in the interval \([1, 2]\), we will analyze the behavior of the function \( f(x) = x^4 - tx + \frac{1}{t} \) on this interval.

1. **Find the derivative of the function:**
   \[
   f(x) = x^4 - tx + \frac{1}{t}
   \]
   \[
   f'(x) = 4x^3 - t
   \]

2. **Set the derivative equal to zero to find critical points:**
   \[
   4x^3 - t = 0
   \]
   \[
   x^3 = \frac{t}{4}
   \]

3. **Determine the range of \( t \) for \( x \) in \([1, 2]\):**
   - For \( x = 1 \):
     \[
     1^3 = \frac{t}{4} \implies t = 4
     \]
   - For \( x = 2 \):
     \[
     2^3 = \frac{t}{4} \implies t = 32
     \]

   Therefore, \( t \) must be in the interval \([4, 32]\) for \( x \) to be in \([1, 2]\).

4. **Check the values of \( t \) given in the options:**
   - \( t = 6 \) is within \([4, 32]\)
   - \( t = 7 \) is within \([4, 32]\)
   - \( t = 8 \) is within \([4, 32]\)
   - \( t = 9 \) is within \([4, 32]\)

   Since all the given values of \( t \) (6, 7, 8, 9) are within the interval \([4, 32]\), the function \( f(x) = x^4 - tx + \frac{1}{t} \) will have a root in the interval \([1, 2]\) for these values of \( t \).

5. **Conclusion:**
   Since none of the given values of \( t \) result in the equation having no root in the interval \([1, 2]\), the correct answer is:
   \[
   \boxed{\text{None of the preceding}}
   \]