1/2 x 1/4 x 1/8 x 1/16 x 1/32 =?
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LM
1
30 tháng 11 2022
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)
HN
2
ND
0
DD
Đoàn Đức Hà
Giáo viên
3 tháng 6 2021
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2\times A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
DD
Đoàn Đức Hà
Giáo viên
3 tháng 6 2021
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2\times B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(B=1-\frac{1}{16}=\frac{15}{16}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4\times x+\frac{15}{16}=1\)
\(\Leftrightarrow4\times x=\frac{1}{16}\)
\(\Leftrightarrow x=\frac{1}{64}\)
HN
0
PN
2
VN
11 tháng 8 2016
Ta có:
1/2+1/4+1/8+1/16+1/32=31/32
Vì có 5 tổng=> ta gọi phần còn lại là 5X
=>5X+31/32=2
=>5X+31/32=64/32
=>5X=64-32-31/32
=>5X=33/32
=>X=33/32:5
=>X=33/160
6 tháng 12 2018
Đề sai nha bạn 
mình sửa luôn
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\dfrac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\dfrac{32}{1-x^{32}}=VP\left(đpcm\right)\)