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2T
3 tháng 9 2019
b) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{64}-1\right)-2^{64}\)
\(=-1\)
S
3 tháng 9 2019
\(\left(1^2-2^2\right)+\left(3^2-4^2\right)+....+\left(99^2-100^2\right)\)
\(=\left(1-2\right)\left(2+1\right)+\left(3-4\right)\left(4+3\right)+....+\left(99-100\right)\left(100+99\right)\)
\(=\left(-1\right)\left(1+2+3+....+100\right)=\frac{\left(-1\right)100.99}{2}=-4950\)
TT
17 tháng 9 2018
A = 12 – 22 + 32 – 42 + … – 20042 + 20052
A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)
A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)
A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005
A = ( 1 + 2002 ). 2005 : 2 = 2011015
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = (22 - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = …
B =(232 - 1)(232 + 1) – 264
B = 264 – 1 – 264
B = - 1
TT
17 tháng 9 2018
xin lỗi nha chỗ câu a mình lộn
chỗ (1+2002)x2005:2=2011015 là sai nha
(1+2005)x2005:2= 2011015 là đúng nha
DP
1
DN
1
T
7 tháng 10 2015
3(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(24-1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(28-1)(28+1)(216+1)(232+1)(264+1)
=(216-1)(216+1)(232+1)(264+1)
=(232-1)(232+1)(264+1)
=(264-1)(264+1)
=(2128-1)
Nếu thấy đúng thì thích cho mình nha
TT
0
PP
2
15 tháng 8 2016
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+2\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\)
15 tháng 8 2016
Ta có ; \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)
= ............................................................................................
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\)
GP
2
10 tháng 8 2016
\(A=-1^2+2^2-3^2+4^2-...-99^2+100^2\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100+99\right)\left(100-99\right)\)
\(=1+2+3+4+...+99+100\)
\(=\frac{\left(1+100\right)\cdot100}{2}=5050\)
\(C=\left(2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=\left(2^{64}-1\right)-2^{42}=-1\)
EC
10 tháng 8 2016
Mk chỉ bt làm câu C thôi tại vì mk chỉ học lớp 7
C=(2+1)(24+1)(28+1)(216+1)(232+1)-264
C=(24-1)(24+1)(28+1)(216+1)(232+1)-264
C=(28-1)(28+1)(216+1)(232+1)-264
C=(216-1)(216+1)(232+1)-264
C=(232-1)(232+1)-264
C=264-1-264
C=-1
G
17 tháng 9 2017
Giải:
1) B = 272 - 252 = (27 - 25)(27 + 25) = 20.52
Suy ra A<B, vì 202<20.52
2) D = 20032 - 1 = 20032 - 12 = (2003 - 1)(2003 + 1) = 2002.2004
Suy ra C = D.
3) Nhân (2-1) vào E, ta đươc: E = (2-1)(2+1)(22+1)(24+1)(28+1)(216+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào E, ta được kế quả:
E = 232-1
Suy ra E<F
4) Nhân (3-1) vào G, ta đươc: 2G = (3-1)(3+1)(32+1)(34+1)(38+1)(316+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào G, ta được kế quả:
2G = 332-1
Suy ra G = (332-1)/2
Mà (332-1)/2 < 332/2
Suy ra G<H
5)
Nhân 2 vào I, ta đươc: 2I = 2.12(52+1)(54+1)(58+1)...(532+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào I, ta được kế quả:
2I = 564-1
Suy ra I = (564-1)/2
Mà (564-1)/2 < 564-1
Suy ra I<K.
Chúc chị học tốt!
D
0
Đặt \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=2^{64}-1\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
Ta có : \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)=2^{64}-1\)
Thay 264 - 1 vào B, ta được :
\(2^{64}-1-2^{64}=-1\)