DN
7
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DN
0
PA
21 tháng 8 2016
\(S=-1^2+2^2-3^2+4^2-...+2016^2\)
\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2016-2015\right)\left(2016+2015\right)\)
\(=3+7+..+4031\)
\(=2033136\)
\(A=\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}\times4^{64}\)
\(15A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^4-1\right)\left(4^4+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^{32}-1\right)\left(4^{32}+1\right)-4^{64}\left(4^{32}\right)\)
\(15A=4^{64}-1-4^{64}\)
\(A=-\frac{1}{15}\)
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)
A
1
DN
30 tháng 8 2018
Ta có:\(A=\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^{32}-1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=4^{64}-1\)
\(\Rightarrow3A=B\)
S
3 tháng 7 2018
Sửa B=432-1
Ta có: \(3A=\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(=\left(4^{16}-1\right)\left(4^{16}+1\right)=4^{32}-1=B\) (đpcm)
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 = \(\frac{-1}{15}\)
Ghi lại cái đề cho rõ hơn đi t giải cho