Let $\dfrac{m}{n} = \dfrac{90^2+5 \times 90+4}{90^2+5\times 90+6} \times \dfrac{91^2+5\times 91+4}{91^2+5\times 91+6} \times .......\times \dfrac{98^2+5\times 98+4}{98^2+5\times 98+6}$ where m and n are co-primes.

What is the value of $|m - n|$ ?

Let the general term be $T(x) = \frac{x^2+5x+4}{x^2+5x+6}$.

Using factorization, $x^2+5x+4 = (x+1)(x+4)$ and $x^2+5x+6 = (x+2)(x+3)$.

So, $T(x) = \frac{(x+1)(x+4)}{(x+2)(x+3)}$.

The given expression is the product of $T(x)$ for $x$ from 90 to 98.

$\frac{m}{n} = \prod_{x=90}^{98} \frac{(x+1)(x+4)}{(x+2)(x+3)}$

$\frac{m}{n} = \frac{(90+1)(90+4)}{(90+2)(90+3)} \times \frac{(91+1)(91+4)}{(91+2)(91+3)} \times ... \times \frac{(98+1)(98+4)}{(98+2)(98+3)}$

$\frac{m}{n} = \frac{91 \times 94}{92 \times 93} \times \frac{92 \times 95}{93 \times 94} \times \frac{93 \times 96}{94 \times 95} \times ... \times \frac{99 \times 102}{100 \times 101}$

Rearranging the terms to see the telescoping product:

$\frac{m}{n} = \left(\frac{91}{92} \times \frac{92}{93} \times \frac{93}{94} \times ... \times \frac{99}{100}\right) \times \left(\frac{94}{93} \times \frac{95}{94} \times \frac{96}{95} \times ... \times \frac{102}{101}\right)$

The first part simplifies to $\frac{91}{100}$.

The second part simplifies to $\frac{102}{93}$.

So, $\frac{m}{n} = \frac{91}{100} \times \frac{102}{93}$.

Simplify the fraction: $\frac{102}{93} = \frac{3 \times 34}{3 \times 31} = \frac{34}{31}$.

$\frac{m}{n} = \frac{91}{100} \times \frac{34}{31} = \frac{91 \times 34}{100 \times 31} = \frac{3106}{3100}$.

To find $m$ and $n$ as co-primes, simplify the fraction $\frac{3106}{3100}$.

Divide numerator and denominator by their greatest common divisor, which is 2.

$\frac{3106 \div 2}{3100 \div 2} = \frac{1553}{1550}$.

1553 and 1550 are co-primes.

Thus, $m = 1553$ and $n = 1550$.

The value of $|m - n| = |1553 - 1550| = 3$.