Đáp án:
\({{3\left( {16{x^2} - 1} \right)} \over {{2^{32}} - 1}}\)
Giải thích các bước giải:
$$\eqalign{
& A = {{16{x^2} - 1} \over {\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)}} \cr
& \left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = {1 \over 3}\left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = {1 \over 3}\left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = {1 \over 3}\left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = {1 \over 3}\left( {{2^{16}} - 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = {1 \over 3}\left( {{2^{32}} - 1} \right) \cr
& \Rightarrow A = {{16{x^2} - 1} \over {{1 \over 3}\left( {{2^{32}} - 1} \right)}} = {{3\left( {16{x^2} - 1} \right)} \over {{2^{32}} - 1}} \cr} $$
