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T
6 tháng 10 2023
\(D=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\) (sửa đề)
\(\dfrac{1}{2}\cdot D=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{4096}\)
\(D-\dfrac{1}{2}D=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{4096}\right)\)
\(\dfrac{1}{2}D=1-\dfrac{1}{4096}\)
\(\dfrac{1}{2}D=\dfrac{4095}{4096}\)
\(\Rightarrow D=\dfrac{4095}{4096}:\dfrac{1}{2}=\dfrac{4095}{2048}\)
Vậy \(D=\dfrac{4095}{2048}\)
BF
0
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 :/
AH
Akai Haruma
Giáo viên
26 tháng 12 2023
Lời giải:
$\frac{3}{4}-(x+1\frac{1}{2})=(-1)^{2024}=1$
$x+\frac{3}{2}=\frac{3}{4}-1=\frac{-1}{4}$
$x=\frac{-1}{4}-\frac{3}{2}=\frac{-7}{4}$
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+1/2^2+1/2^3+.....+1/2^10)
đặt A=(1/2+1/2^2+1/2^3+.....+1/2^10)
2A=2(1/2+1/2^2+1/2^3+.....+1/2^10)=1+1/2+...+1/2^9
A=(1+1/2+...+1/2^9)-(1/2+...+1/2^10)
A=1-1/2^10
=-1-1-1/2^10=......tự làm nha
Đề chắc sai e ạ, a sửa luôn :
\(A=\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)
\(A=\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(2A=1-\frac{1}{2}-...-\frac{1}{2^9}\)
\(2A-A=\left(1-\frac{1}{2}-...-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2}-...-\frac{1}{2^9}-\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(A=1-\left(\frac{1}{2}+\frac{1}{2}\right)+\frac{1}{2^{10}}\)
\(A=\frac{1}{2^{10}}\)