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=\(1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{10\cdot12}\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)
=1+12⋅4+14⋅6+...+110⋅121+12⋅4+14⋅6+...+110⋅12
=1+12(12−14+14−16+...+110−112)=1+12(12−14+14−16+...+110−112)
=1+12⋅512=1+524=2924
\(=\dfrac{1}{8}\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)\)
\(=\dfrac{1}{8}\cdot2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=\dfrac{1}{4}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{6}{7}=\dfrac{3}{14}\)
A= 1/3+1/6+1/12+1/24+1/48+1/96
= (1/3+1/6)+(1/12+1/24)+(1/48+1/96)
= (2/6+1/6)+(2/24+1/24)+(2/96+1/96)
= 1/2+1/8+1/32
= 16/32+4/32+1/32
= 21/32
Vậy A=21/32
Giải:
A=1/3+1/6+1/12+1/24+1/48+1/96
A=1/3+(1/2.3+1/3.4)+(1/4.6+1/6.8)+1/96
A=1/3+(1/2-1/3+1/3-1/4)+[1/2.(2/4.6+2/6.8)]+1/96
A=1/3+(1/2-1/4)+[1/2.(1/4-1/6+1/6-1/8)]+1/96
A=1/3+1/4+[1/2.(1/4-1/8)]+1/96
A=1/3+1/4+[1/2.1/8]+1/96
A=1/3+1/4+1/16+1/96
A=7/12+7/96
A=21/32
câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)
\(B=\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\left(1+\dfrac{1}{24}\right).....\left(1+\dfrac{1}{440}\right)\left(1+\dfrac{1}{483}\right)\)
\(B=\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.....\dfrac{441}{440}.\dfrac{484}{483}\)
\(B=\dfrac{9.16.25.....441.484}{8.15.24.....440.483}\)
\(B=\dfrac{3.3.4.4.5.5.....21.21.22.22}{2.4.3.5.4.6.....20.22.21.23}\)
\(B=\dfrac{3.4.5.....21.22}{2.3.4.....20.21}.\dfrac{3.4.5.....21.22}{4.5.6.....22.23}\)
\(B=11.\dfrac{3}{23}=\dfrac{33}{23}\)
B = \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{121}{120}.\dfrac{144}{143}\)
B = \(\dfrac{4.9.16.25...121.144}{3.8.15.24....120.143}\)
B = \(\dfrac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)
B = \(\dfrac{2.3.4.5...11.12}{1.2.3.4.5...10.11}.\dfrac{2.3.4.5...11.12}{3.4.5.6.7...12.13}\)
B = 12 . \(\dfrac{2}{13}\)
B = \(\dfrac{24}{13}\)
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{30}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{29}{30}\)
\(=\dfrac{1.2.3...29}{2.3.4...30}\)
\(=\dfrac{1}{30}\)
\(B=1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.1\dfrac{1}{24}...1\dfrac{1}{168}\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{169}{168}\)
\(=\dfrac{4.9.16.25...169}{3.8.15.24...168}\)
\(=\dfrac{2.3.4...13}{1.2.3...12}.\dfrac{2.3.4...13}{3.4.5...14}\)
\(=13.\dfrac{1}{7}\)
\(=\dfrac{13}{7}\).
A = \(\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{29}{30}=1x1x1x...x\dfrac{1}{30}=\dfrac{1}{30}\)
B = \(\dfrac{4}{3}x\dfrac{9}{8}x\dfrac{16}{15}x\dfrac{25}{24}x...x\dfrac{169}{168}=1x1x1x1x...x\dfrac{13}{7}=\dfrac{13}{7}\)
Câu B em chưa rõ cách làm nhanh cho lắm. Nếu ko cần tính nhanh thì chị có thể giải bình thường ra giấy ha.
Ta có: \(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(\Leftrightarrow2A=2+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=2+\dfrac{1}{2}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{24}{12}+\dfrac{6}{12}-\dfrac{1}{12}\)
\(\Leftrightarrow2A=\dfrac{29}{12}\)
hay \(A=\dfrac{29}{24}\)