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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 \(A=\dfrac{29}{24}\)
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
=\(1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{10\cdot12}\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)
=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
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{30}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{29}{30}\)
\(=\dfrac{1.2.3...29}{2.3.4...30}\)
\(=\dfrac{1}{30}\)
\(B=1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.1\dfrac{1}{24}...1\dfrac{1}{168}\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{169}{168}\)
\(=\dfrac{4.9.16.25...169}{3.8.15.24...168}\)
\(=\dfrac{2.3.4...13}{1.2.3...12}.\dfrac{2.3.4...13}{3.4.5...14}\)
\(=13.\dfrac{1}{7}\)
\(=\dfrac{13}{7}\).
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.
\(\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}-\dfrac{47}{24}\right)+\dfrac{2}{3}=\dfrac{37}{37}-\dfrac{13}{8}+\dfrac{2}{3}=\dfrac{1}{24}\)
Bài 2 :
a, \(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-30}{40}=-\dfrac{6}{40}=-\dfrac{3}{20}\)
b, \(2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)
\(=\dfrac{1}{8}\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)\)
\(=\dfrac{1}{8}\cdot2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=\dfrac{1}{4}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{6}{7}=\dfrac{3}{14}\)