\(=\dfrac{3.4-\dfrac{3.4}{4}-\dfrac{3.4}{289}-\dfrac{3.4}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}=\dfrac{3.\left(4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}\right)}{1.\left(4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}\right)}=\dfrac{3}{1}=3\)