Đáp án:
Giải thích các bước giải:
Áp dụng \[\left( {x - 1} \right)\left( {x + 1} \right) = {x^2} - 1\]
Ta có
\[\begin{array}{l}
3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right).....\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right).....\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{{\left( {{2^2}} \right)}^2} - 1} \right)\left( {{2^4} + 1} \right)......\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)......\left( {{2^{64}} + 1} \right) + 1\\
= \left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)......\left( {{2^{64}} + 1} \right) + 1\\
.....\\
= \left( {{2^{64}} - 1} \right)\left( {{2^{64}} + 1} \right) + 1\\
= {2^{128}} - 1 + 1 = {2^{128}}
\end{array}\]
