Tính A=(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)-1/15 . 4^64
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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 = \(\frac{-1}{15}\)
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:\(A=\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^{32}-1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=4^{64}-1\)
\(\Rightarrow3A=B\)
BV
24 tháng 6 2014
Tính không quy đồng mẫu:
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{64}\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
3 tháng 8 2023
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(\dfrac{4}{2}A=\dfrac{4}{2}\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+..\left(\dfrac{1}{32}-\dfrac{1}{32}\right)+\left(1-\dfrac{1}{64}\right)\)
\(A=1-\dfrac{1}{64}\)
\(A=\dfrac{63}{64}\)
PA
21 tháng 8 2016
\(S=-1^2+2^2-3^2+4^2-...+2016^2\)
\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2016-2015\right)\left(2016+2015\right)\)
\(=3+7+..+4031\)
\(=2033136\)
\(A=\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}\times4^{64}\)
\(15A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^4-1\right)\left(4^4+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^{32}-1\right)\left(4^{32}+1\right)-4^{64}\left(4^{32}\right)\)
\(15A=4^{64}-1-4^{64}\)
\(A=-\frac{1}{15}\)
LM
2 tháng 5 2022
\(\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{8}{64}+\dfrac{4}{64}+\dfrac{2}{64}+\dfrac{1}{64}=\dfrac{32+16+8+4+2+1}{64}=\dfrac{63}{64}\)
