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1/5 + 1/5 - 1/10 + 1/10 - 1/20 + 1/20 - 1/40 + ... + 1/640 - 1/1280
= 1/5 + 1/5 - 1/1280 = 511/1280
B=51+101+201+401+...+12801
�=1⋅15+12⋅15+14⋅15+18⋅15+...+1256⋅15B=1⋅51+21⋅51+41⋅51+81⋅51+...+2561⋅51
�=15⋅(1+12+14+18+...+1256)B=51⋅(1+21+41+81+...+2561)
Đặt �=1+12+14+18+...+1256A=1+21+41+81+...+2561
⇒2�=2+1+12+14+...+1128⇒2A=2+1+21+41+...+1281
⇒2�−�=2−1256⇒2A−A=2−2561
�=2−1256A=2−2561
Thay A vào B
có: �=15.(2−1256)=15⋅511256=5111280B=51.(2−2561)=51⋅256511=1280511
hướng dẫn em nhé:
quy đồng mẫu số lên có mẫu số chung là 160 rồi cộng lại em nhé
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
= \(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\)
=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
=\(\frac{1}{3}-\frac{1}{7}\)
=\(\frac{4}{21}\)
A = \(\dfrac{1}{5}+\dfrac{1}{10}+...+\dfrac{1}{1280}\)
= \(\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
= \(\dfrac{2}{5}-\dfrac{1}{1280}=\dfrac{511}{1280}\)
Giải:
\(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...+\dfrac{1}{1280}\)
\(=\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{40}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
\(=\dfrac{2}{5}-\dfrac{1}{1280}\)
\(=\dfrac{511}{1280}\)
\(=\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{40}+...+\dfrac{1}{640}-\dfrac{1}{1280}\\ =\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{1280}\\ =\dfrac{2}{5}-\dfrac{1}{1280}\\ =\dfrac{511}{1280}\)
=2/5 nha
cho mik xin 1 tick mik cảm ơn
1/5 + 1/10 + 1/20 + 1/40 + 1/40
= 3/10 + 1/20 + 1/40 + 1/40
= 7/20 + 1/40 + 1/40
= 3/8 + 1/40
= 2/5