LN
2
K
Khách
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DN
0
KT
30 tháng 10 2018
a) 153^2+99.153+47^2
= 153^2+2.47.153+47^2
= (153+47)^2
=200^2
=40000
b) 126^2-152.126+5776
= 126^2-2.76.126+76^2
= (126-76)^2
= 50^2
= 2500
c)3^8.5^8-(15^4-1).(15^4+1)
= 15^8-[(15^4)^2-1^2]
= 15^8-15^8+1
=1
d) (2+1).(2^2+1).(2^4+1)...(2^32+1)+1
= 1.(2+1).(2^2+1).(2^4+1)...(2^32+1)+1
= (2-1).(2+1).(2+1).(2^4+1)...(2^32+1)+1
= (2^2-1).(2^2+1).(2^4+1)...(2^32+1)+1
= (2^4-1).(2^4+1)...(2^32+1)+1
= (2^8-1)...(2^32+1)+1
= (2^32-1).(2^32+1)+1
= 2^64-1+1
= 2^64
DN
7
27 tháng 8 2016
Ta có (42 - 1)(42 + 1) = 44 - 1
Ta có 15A = (42 - 1)(42 + 1)(44 + 1)(48 + 1)(416 + 1)(432 + 1) - 464 = 464 - 1 - 464 = -1
=> A = \(\frac{-1}{15}\)
10 tháng 8 2023
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
10 tháng 8 2023
8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
ND
0
TH
16 tháng 8 2019
Bài 2:
Ta có: \(\frac{x+1}{x}=10\) hay \(\frac{x^1+1^1}{x^1}=10^1\)
Nên suy ra : \(\frac{x^5+1}{x^5}=10^5\)
= 100000 ( do 15 cũng sẽ =1 nên không viết mũ 5 cũng chả sao)
HV
4
16 tháng 7 2019
38.58−(154−1)(154+1)38.58−(154−1)(154+1)
=158−(158−1)=158−(158−1)
=158−158+1=1
Học tốt
KN
16 tháng 7 2019
\(3^8.5^8-\left(15^4-1\right)\left(15^4+1\right)\)
\(=15^8-\left[\left(15^4\right)^2-1^2\right]\)
\(=15^8-\left[15^8-1\right]\)
\(=15^8-15^8+1=1\)
TN
1
15 tháng 7 2016
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow2^{64}-1\)
đặt
\(A=15.\left(4^2+1\right)\left(4^4+1\right)..\left(4^{32}+1\right)\)
ta có
\(A=\left(4^2-1\right)\left(4^2+1\right)..\left(4^{32}+1\right)\)
\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)..\left(4^{32}+1\right)=..=4^{64}-1\)
Đặt \(A=15\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)
\(=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)
\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)
\(=\left(4^{32}-1\right)\left(4^{32}+1\right)=4^{64}-1\)