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\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-...-\frac{1}{11.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=1-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{57}{130}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}-\frac{1}{3}-\frac{1}{15}-.....-\frac{1}{143}\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{143}\right)\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{11.13}\right)\)
\(=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)\(=\left(\frac{1}{1}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{117}{130}-\frac{78}{130}=\frac{39}{130}=\frac{3}{10}\)
\(\frac{1}{\left(k-1\right)k\left(k+1\right)}=\frac{k+1-\left(k-1\right)}{\left(k-1\right)k\left(k+1\right)}\frac{1}{2}\)\(=\frac{1}{2}\left(\frac{k+1}{\left(k-1\right)k\left(k+1\right)}-\frac{k-1}{\left(k-1\right)k\left(k+1\right)}\right)\)\(=\frac{1}{2}\left(\frac{1}{\left(k-1\right)k}-\frac{1}{k\left(k+1\right)}\right)\)
\(\Rightarrow\)VT\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
ta có
(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)
=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)
19/36+17/90+1/12
=(19/36+1/12)+17/90
=7/12+17/90
=105/180+34/180
=139/180
1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45
=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5
=1/1-1/3+1/3-1/2....+1/9-1/5
=1/1
Tự chép đề
\(A=\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{8\cdot9}+\frac{2}{9\cdot11}+\frac{3}{11\cdot14}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{7}{14}-\frac{1}{14}=\frac{6}{14}=\frac{3}{7}\)
Vậy \(A=\frac{3}{7}\)
bn ơi bài này ko có 1/4 đâu:
đặt A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{36}+\)\(\frac{1}{45}\)
\(\frac{1}{2}A\)=> \(\frac{1}{2}A\)= \(\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)
\(\frac{1}{2}A\)= \(\frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(\frac{1}{2}A\)= \(\frac{1}{4}\)+ \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}A\)= \(\frac{1}{4}+\frac{1}{2}-\frac{1}{10}=\frac{5}{20}+\frac{10}{20}-\frac{2}{20}=\frac{13}{20}\)
=> A = \(\frac{13}{20}:\frac{1}{2}=\frac{13}{10}\)
Chúc bn học tốt !
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Hỏi một đằng trả lời một nẻo là sao???