(i) \(\frac{1351}{1250}\) = \(\frac{1351}{5^4\times2}\)

We know 2 and 5 are not the factors of 1351. 

So, the given rational is in its simplest form. 

And it is of the form (2^m × 5ⁿ) for some integers m, n. 

So, the given number is a terminating decimal.

∴ \(\frac{1351}{5^4\times2}\) = \(\frac{1351\times2}{5^4\times2^4}\) = \(\frac{10808}{10000}\) = 1.0808

(ii) \(\frac{2017}{250}\) = \(\frac{2017}{5^3\times2}\)

We know 2 and 5 are not the factors of 2017. 

So, the given rational is in its simplest form. 

And it is of the form (2^m × 5ⁿ) for some integers m, n. 

So, the given rational number is a terminating decimal.

∴ \(\frac{2017}{5^3\times2}\) = \(\frac{2017\times2^2}{5^3\times2^3}\) = \(\frac{8068}{1000}\) = 8.068

(iii) \(\frac{3219}{1800}\) = \(\frac{3219}{2^3\times5^2\times3^2}\)

We know 2, 3 and 5 are not the factors of 3219. 

So, the given rational is in its simplest form. 

∴ (2³ × 5² × 3² ) ≠ (2^m × 5ⁿ) for some integers m, n.

Hence, \(\frac{3219}{1800}\) is not a terminating decimal.

\(\frac{3219}{1800}\) = 1.78833333….

Thus, it is a repeating decimal.

\(\frac{1723}{625}\) = \(\frac{1723}{5^4}\)

We know 5 is not a factor of 1723. 

So, the given rational number is in its simplest form. 

And it is not of the form (2^m × 5ⁿ)

Hence, \(\frac{1723}{625}\)is not a terminating decimal.