NP
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K
Khách
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HH
26 tháng 6 2018
Giải:
a) \(M=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=2^{32}-1\)
\(\Leftrightarrow M=\dfrac{2^{32}-1}{3}\)
Vậy ...
b) \(N=16\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^{16}-1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=7^{32}-1\)
\(\Leftrightarrow N=\dfrac{7^{32}-1}{3}\)
Vậy ...
TN
8 tháng 6 2016
B=3(22 +1)(24+1)(28+1)(216+1)
=(4-1)(22+1)(24+1)(28+1)(216+1)
=[(22-1)(22+1)](24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)
=(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
9 tháng 6 2016
B=3(22 +1)(24+1)(28+1)(216+1)
=(4-1)(22+1)(24+1)(28+1)(216+1)
=[(22-1)(22+1)](24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)
=(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
DN
1 tháng 9 2018
đặt biểu thức \(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\) là A
Ta có:\(A=\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^{16}-1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=4^{32}-1\)
\(\Rightarrow A=\dfrac{4^{32}-1}{15}\)
Vậy giá trị biểu thức trên là \(\dfrac{4^{32}-1}{15}\)
\(b,40^2-39^2+38^2-37^2+...+2^2-1^2\)
\(=\left(40^2-39^2\right)+\left(38^2-37^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(40-39\right)\left(40+39\right)+\left(38-37\right)\left(38+37\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=1+2+...+38+39+40\)
\(=\dfrac{\left(40+1\right).40}{2}=\dfrac{41.40}{2}=820\)
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
7 tháng 10 2017
\(100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=1.199+1.195+...+1.3\)
\(=199+195+....+3\)
\(=\left[\left(\dfrac{199-3}{4}\right)+1\right]:2.\left(199+3\right)=5050\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
\(3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)
\(=\left(2^8-1\right)......\left(2^{64}+1\right)=2^{128}-1\)
3 tháng 7 2015
tách ít ít ra thôi. để cả cộp thế này k ai làm cho đâu. mệt quá
HM
29 tháng 5 2018
\(A=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(A=2^2-1^2+4^2-3^2+...+100^2-99^2\)
\(A=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)
\(A=1\left(1+2\right)+1\left(3+4\right)+....+1\left(99+100\right)\)
\(A=1+2+3+4+....+99+100\)
A=5050
\(B=3^8.7^8-\left(21^4-1\right)\left(21^4+1\right)\)
\(B=\left(3.7\right)^8-\left(21^8-1\right)\)
\(B=21^8-21^8+1\)
B=1
mà A=5050
⇒ A>B
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
\(\left(1^2-2^2\right)+\left(3^2-4^2\right)+....+\left(99^2-100^2\right)\)
\(=\left(1-2\right)\left(2+1\right)+\left(3-4\right)\left(4+3\right)+....+\left(99-100\right)\left(100+99\right)\)
\(=\left(-1\right)\left(1+2+3+....+100\right)=\frac{\left(-1\right)100.99}{2}=-4950\)
PP
2
HN
4 tháng 7 2018
Mình làm câu c trước để bạn hình dung ra nhé, câu a tương tự:
c) \(7\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(8-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left[\left(2^3-1\right)\left(2^3+1\right)\right]\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^6-1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^{12}-1\right)\left(2^{24}+1\right)\)
\(=2^{36}-1\)
b) \(\left(x^2-x+4\right)\left(x^2+x+1\right)\left(x^2-1\right)\)
\(=\left(x^2.x^2.x^2\right).\left(-x+4+x+1+\left(-1\right)\right)\)
\(=x^8.\left(-4\right)\)
4 tháng 7 2018
\(a,\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\)
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\)