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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. \(3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\)
a. \({\left( {6x + 1} \right)^2} + {\left( {6x – 1} \right)^2} – 2\left( {1 + 6x} \right)\left( {6x – 1} \right)\)
Advertisements (Quảng cáo)
\(\eqalign{ & = {\left( {6x + 1} \right)^2} – 2\left( {6x + 1} \right)\left( {6x – 1} \right) + {\left( {6x – 1} \right)^2} = {\left[ {\left( {6x + 1} \right) – \left( {6x – 1} \right)} \right]^2} \cr & = {\left( {6x + 1 – 6x + 1} \right)^2} = {2^2} = 4 \cr} \)
b. \(3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\)
\(\eqalign{ & = \left( {{2^2} – 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr & = \left( {{2^4} – 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) = \left( {{2^8} – 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr & = \left( {{2^{16}} – 1} \right)\left( {{2^{16}} + 1} \right) = {2^{32}} – 1 \cr} \)