Nội dung bài giảng
Rút gọn biểu thức
a. \({\left( {6x + 1} \right)^2} + {\left( {6x - 1} \right)^2} - 2\left( {1 + 6x} \right)\left( {6x - 1} \right)\)
b. \(3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\)
Giải:
a. \({\left( {6x + 1} \right)^2} + {\left( {6x - 1} \right)^2} - 2\left( {1 + 6x} \right)\left( {6x - 1} \right)\)
\(\eqalign{ & = {\left( {6x + 1} \right)^2} - 2\left( {6x + 1} \right)\left( {6x - 1} \right) + {\left( {6x - 1} \right)^2} = {\left[ {\left( {6x + 1} \right) - \left( {6x - 1} \right)} \right]^2} \cr & = {\left( {6x + 1 - 6x + 1} \right)^2} = {2^2} = 4 \cr} \)
b. \(3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\)
\(\eqalign{ & = \left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr & = \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) = \left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr & = \left( {{2^{16}} - 1} \right)\left( {{2^{16}} + 1} \right) = {2^{32}} - 1 \cr} \)