Question

Rút gọn biểu thức

3(2^2 +1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)

Answer 1

Ta có : $3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$

$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)$

$=(2^{16}-1)(2^{16}+1)(2^{32}+1)$

$=(2^{32}-1)(2^{32}+1)$

$=2^{64}-1$

Answer 2

3.(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)

= (2²-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)

= (2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)

= (2⁸-1)(2⁸+1)(2¹⁶+1)(2³²+1)

= (2¹⁶-1)(2¹⁶+1)(2³²+1)

= (2³²-1)(2³²+1)

= 2⁶⁴ - 1

Gợi ý: Áp dụng hằng đẳng thức số 3 trong 7 hằng đẳng thức đáng nhớ