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1) \(1+\frac{1}{3}+1\frac{91}{27}+\frac{1}{81}\)
\(=1+\frac{1}{3}+4+\frac{10}{27}+\frac{1}{81}\)
\(=\left(1+4\right)+\left(\frac{1}{3}+\frac{10}{27}+\frac{1}{81}\right)\)
\(=5+\frac{58}{81}=5\frac{58}{81}\)
2) \(\frac{1}{78}+\frac{1}{91}+\frac{1}{105}+\frac{1}{120}+\frac{1}{136}\)
\(=\frac{1}{6\times13}+\frac{1}{13\times7}+\frac{1}{7\times15}+\frac{1}{15\times8}+\frac{1}{8\times17}\)
\(=\frac{1}{6}-\frac{1}{13}+\frac{1}{13}-\frac{1}{7}+\frac{1}{7}-\frac{1}{15}+\frac{1}{15}-\frac{1}{8}+\frac{1}{8}-\frac{1}{17}\)
(gạch bỏ các số giống nhau)
\(=\frac{1}{6}-\frac{1}{17}=\frac{11}{102}\)
3) \(\frac{1}{6}+\frac{1}{30}+\frac{1}{70}+\frac{1}{126}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times10}+\frac{1}{10\times7}+\frac{1}{7\times18}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{7}+\frac{1}{7}-\frac{1}{18}\)
\(=\frac{1}{2}-\frac{1}{18}=\frac{4}{9}\)
chúc bn học tốt!!
công chúa ôri ơi mk viết nhầm đề bài 1
đề bài là 1+1/3+1/9+1/27+1/81
bạn trả lời giúp mk với mk cần gấp lắm
cảm ơn
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
= \(\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{12}+\frac{1}{24}\right)+\left(\frac{1}{48}+\frac{1}{96}\right)+\frac{1}{192}\)
= \(\left(\frac{1}{2}+\frac{1}{8}\right)+\left(\frac{1}{32}+\frac{1}{192}\right)\)
= \(\frac{5}{8}+\frac{1}{192}\)
= \(\frac{121}{192}\)
94.2023+2023:1/6
=94.2023+2023.6
=(94+6).2023
=100.2023
=202300
A=1.4+1/4.7+1/7.10+...+1/91.94
=1/3.(3/1.4+3/4.7+3/7.10+...+3/91.94)
=1/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/91-1/94)
=1/3.(1-1-94)
=1/3.(93/94)
=31/94
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{91\cdot94}\)
\(=\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{91\cdot94}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{94}\right)\)
\(=\frac{1}{3}\cdot\frac{93}{94}\)
\(=\frac{31}{94}\)