First, the exact mes of the nuclei are not the same as the m numbers in the equations. There are very slight variations in m, but these variations are exceedingly important. In order to calculate the actual loss of m, you would need the mes of the various nuclei and neutrons to several decimal places. Then, when you add up all the mes, you see that there is a very small change in the mass during the reaction. Since c^2 is such a huge number, even very small mass changes result in the release of a lot of energy.
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