1- 1/2 -1/4 -1/8 -1/16
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11 tháng 7 2021
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
DT
2
12 tháng 12 2020
Bài làm
\(\frac{\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)}{\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)}\)
\(=\frac{\left(\frac{2}{2}+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\right)}{\left(\frac{2}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}\right)}\)
\(=\frac{\frac{1}{2}\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\right)}{\frac{1}{2}\left(2-1+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}\right)}\)
\(=\frac{3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}}{1+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}}\)
\(=\frac{\frac{24}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}}{\frac{8}{8}+\frac{4}{8}-\frac{2}{8}+\frac{1}{8}}\)
\(=\frac{31}{8}\div\frac{11}{8}\)
\(=\frac{31}{8}\cdot\frac{8}{11}\)
\(=\frac{31}{11}\)
P/S: Trông không thuận tiện lắm :/
PT
2
2 tháng 9 2015
\(\frac{\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)}{\left(1-\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)}=\frac{\left(\frac{16}{16}+\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)}{\left(\frac{16}{16}-\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)}=\frac{\frac{31}{16}}{\frac{15}{16}}=\frac{31}{16}:\frac{15}{16}=\frac{31}{16}\times\frac{16}{15}=\frac{31}{15}\)
BT
1
BT
1
13 tháng 8 2018
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
\(=\left(2^4\right)^8-1=16^8-1\)
\(\Rightarrow D=1\)Chúc bạn học tốt.
HP
1
17 tháng 12 2022
\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+...+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+...+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)
HH
0
PD
1
H9
HT.Phong (9A5)
CTVHS
18 tháng 10 2023
\(1+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{5}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{11}{8}+\dfrac{1}{16}\)
\(=\dfrac{23}{16}\)
______
\(2-\dfrac{1}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{15}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{43}{24}-\dfrac{1}{16}\)
\(=\dfrac{83}{48}\)
_________
\(\dfrac{4}{99}\times\dfrac{18}{5}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}\times\dfrac{11}{12}+\dfrac{3}{5}\)
\(=\dfrac{2}{15}+\dfrac{3}{5}\)
\(=\dfrac{11}{15}\)
__________
\(\left(1-\dfrac{3}{4}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{4}\times\dfrac{4}{3}\times\dfrac{2}{3}\)
\(=\dfrac{4\times2}{4\times3\times3}\)
\(=\dfrac{2}{3\times3}\)
\(=\dfrac{2}{9}\)
DN
1
23 tháng 7 2016
\(A=\frac{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}}{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}}\)
Đặt tử số là B, mẫu số là C
\(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(2B-B=\left(2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)\)
\(B=2-\frac{1}{16}\)
\(B=\frac{32}{16}-\frac{1}{16}=\frac{31}{16}\)
\(C=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\)
\(2C=2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\)
\(2C+C=\left(2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)\)
\(3C=2+\frac{1}{16}\)
\(3C=\frac{32}{16}+\frac{1}{16}\)
\(3C=\frac{33}{16}\)
\(C=\frac{33}{16}:3=\frac{11}{16}\)
=> \(A=\frac{B}{C}=\frac{31}{16}:\frac{11}{16}=\frac{31}{16}.\frac{16}{11}=\frac{31}{11}\)
1-1/2-1/4-1/8-1/16=1/16