Bài 56 trang 14 SBT toán 8 tập 1

Đề bài

Rút gọn biểu thức

\(a)\) \({\left( {6x + 1} \right)^2} + {\left( {6x - 1} \right)^2}\)\( - 2\left( {1 + 6x} \right)\left( {6x - 1} \right)\)

\(b)\)\(3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\)

Lời giải

\(a)\) \({\left( {6x + 1} \right)^2} + {\left( {6x - 1} \right)^2}\)\( - 2\left( {1 + 6x} \right)\left( {6x - 1} \right)\)

\(= {\left( {6x + 1} \right)^2} - 2\left( {6x + 1} \right)\left( {6x - 1} \right) \)\(+ {\left( {6x - 1} \right)^2}\)

\( = {\left[ {\left( {6x + 1} \right) - \left( {6x - 1} \right)} \right]^2} \)

\( = {\left( {6x + 1 - 6x + 1} \right)^2} = {2^2} = 4 \)

\(b)\) \(3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\)

\( = \left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\)

\(  = \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \)

\( = \left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\)

\(= \left( {{2^{16}} - 1} \right)\left( {{2^{16}} + 1} \right) = {2^{32}} - 1  \)


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