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a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
a) (2+1)(2^2+1)(2^4+1)...(2^32+1)-2^64
=(2+1)(2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64
=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64
=(2^4-1)(2^4+1)....(2^32+1)-2^64
=......
=(2^32-1)(2^32+1)-2^64
=2^64-1-2^64=-1
b)Đặt A=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)+(5^128-3^128)/2
đặt B=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)
\(2B=\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)
\(2B=\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)
\(2B=\left(5^4-3^4\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)
\(2B=.......\)
2B=(5^64-3^64)(5^64+3^64)
2B=5^128-3^128
B=(5^128-3^128)/2 (thế vào đề bài)
=> A=B+(5^128-3^128)/2=(5^128-3^128)/2+(5^128-3^128)/2=\(\frac{2\left(5^{128}-3^{128}\right)}{2}=\left(5^{128}-3^{128}\right)\)
a) A = ( 2-1)(2+1)(22+1)...(232+1)-264
=(22-1)(22+1)(24+1)... -264
=....
=264-1-264=1
câu b tương tự nhá
a, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=2^{64}-1-2^{64}=-1\)
b,\(B=\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)
\(=\dfrac{\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)}{2}+\dfrac{5^{128}-3^{128}}{2}\)\(=\dfrac{\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}\)
\(=\dfrac{\left(5^{64}-3^{64}\right)\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}=\dfrac{2.5^{128}}{2}=5^{128}\)
Ta có \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^{64}-1\right)\left(3^{64}+1\right)=\left(3^{128}-1\right)\)
\(\Rightarrow A=\frac{3^{128}-1}{2}\)
3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
4:
D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)
=(4^8-1)(4^8+1)*...*(4^64+1)
=...
=4^128-1
5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)
=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1
=5^256-1+5^256-1
=2*5^256-2
A = 1002 - 992 + 982 - 972 + . . . + 22 - 12
= (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + . . . (2 - 1)(2 + 1)
= 199 + 195 + . . . + 3
= 5050
B = 3(22 + 1)(24 + 1) . . . (264 + 1) + 1
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1)(264 + 1) + 1
= (24 - 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (28 - 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (216 - 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (232 - 1)(232 + 1)(264 + 1) + 1
= (264 - 1)(264 + 1) + 1
= 2128 - 1 + 1
= 2128
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
Giúp vs nhé mk sẽ dùng các nick phụ tới tấp cho mí bạn