\(\frac{5}{8}\)= \(\frac{20}{32}\)= \(\frac{80}{128}\)= ?
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Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
\(=\frac{1}{1x2}+\frac{1}{2x4}+\frac{1}{4x8} +\frac{1}{8x16}+\frac{1}{16x32}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
nè ..!
\(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}\)
\(=\frac{256+64+16+4+1}{512}\)
\(=\frac{341}{512}\)
Bài làm:
Đặt \(A=\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}=\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^8}\)
=> \(2A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
=> \(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^8}\right)\)
<=> \(A=\frac{1}{2^2}-\frac{1}{2^8}=\frac{2^6-1}{2^8}=\frac{64-1}{256}=\frac{63}{256}\)
đáp án là 63/256 nha bạn ~ mk hơi bận nên k kịp trình bày!
= 1 - 1/2+ 1/2- 1/4 +1/4 - 1/8 +1/8 -1/16 +1/16 -1/32 +1/32 -1/64 +1/64 - 1/128
= 1-1/128
=127/128
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{64}{128}\)+ \(\frac{32}{128}\)+ \(\frac{16}{128}\)+ \(\frac{8}{128}\)+ \(\frac{4}{128}\)+ \(\frac{2}{128}\)+ \(\frac{1}{128}\).
= \(\frac{127}{128}\).
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)
= \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)+ \(\frac{1}{32}\)- \(\frac{1}{64}\)+ \(\frac{1}{64}\)- \(\frac{1}{128}\)
= \(1\)- \(\frac{1}{128}\)
= \(\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
=\(1-\frac{1}{256}\)
=\(\frac{255}{256}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
= 128/256 + 64/256 + 32/256 + 16/256 + 8/256 + 4/256 + 2/128 + 1/256
= 255/256
)
58 = 2032 = 80128 = \(\frac{320}{512}\)
5 x 4 = 20 ; 20 x 4 = 80
=> 80x 4 = 320
128 x 4 = 512
\(\frac{320}{512}\)