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\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
dùng hằng đẳng thức A^2 - B^2 = (A - B)(A + B) nhé phần b chuyển vế sang rồi dùng hđt là Okay
Nhân với 2-1 áp dụng bất đẳng thức a^2-b^2=(a-b)(a+b)
=> 2^64-1
(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)
=[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)
(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (24 - 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (28 - 1)(28 + 1)(216 + 1) - 232
= (216 - 1)(216 + 1) - 232
= (232 - 1) - 232
= 232 - 1 - 232
= -1
Ta có :
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\left(đpcm\right)\)
\(3\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)\)
\(=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)\)
\(=\left(2^8-1\right).\left(2^8+1\right)\)
\(=2^{16}-1\)
Vậy \(3.\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)=2^{16}-1\left(đpcm\right)\)
\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
b/ \(100^2+\left(100+3\right)^2+\left(100+5\right)^2+\left(100-6\right)^2\)
\(=100^2+100^2+100^2+100^2+4.100+9+25+36\)
\(=100^2+2.100+1+100^2-4.100+4+100^2-8.100+16+100^2+14.100+49\)
\(=\left(100+1\right)^2+\left(100-2\right)^2+\left(100-4\right)^2+\left(100+7\right)^2\)
(a + b)(a2 + b2)(a4 + b4)(a8 + b8)(a16 + b16)
=1.(a + b)(a2 + b2)(a4 + b4)(a8 + b8)(a16 + b16)
= (a – b) (a + b)(a2 + b2)(a4 + b4)(a8 + b8)(a16 + b16)
= (a2 – b2) (a2 + b2)(a4 + b4)(a8 + b8)(a16 + b16)
= (a4 – b4)(a4 + b4)(a8 + b8)(a16 + b16)
= (a8 – b8)(a8 + b8)(a16 + b16)
= (a16– b16)(a16 + b16)
= a32 – b32
Giúp mình làm bài này nhé!!!
Ta có:
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Vậy...