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Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)
\(A=\frac{1}{5}-\frac{1}{605}\)
\(A=\frac{24}{121}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{602\cdot605}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)
\(=\frac{1}{5}-\frac{1}{605}+0+...+0\)
\(=\frac{24}{121}\)
\(B=\frac{5}{2}+\left(\frac{4}{1.11}+\frac{3}{11.2}\right)+\left(\frac{1}{2.15}+\frac{13}{15.4}\right)\)
\(B=\frac{5}{2}+\frac{1}{11}.\left(4+\frac{3}{2}\right)+\frac{1}{15}\left(\frac{1}{2}+\frac{13}{4}\right)=\frac{5}{2}+\frac{1}{11}.\frac{11}{2}+\frac{1}{15}.\frac{15}{4}\)
=> \(B=\frac{5}{2}+\frac{1}{2}+\frac{1}{4}=\frac{10}{4}+\frac{2}{4}+\frac{1}{4}=\frac{13}{4}\)
Ta có
1/7.B = 5/2.7 + 4/7.11 + 3/11.14 + 1/14.15 + 13/15.28
1/7.B = 1/2 - 1/7 + 1/7 - 1/11 + 1/11 - 1/14 + 1/14 - 1/15 + 1/15 - 1/28
1/7.B = 1/2 - 1/28
1/7.B = 14/28 - 1/28
1/7.B = 13/28
B = 13/28 : 1/7
B = 13/28 . 7
B = 13/4
B = \(\left(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\right)\))
B = 7 . \(\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
B = 7 . \(\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
B = 7 . ( 1/2 - 1/28 )
B = 7 . 13/28
B = 13/4
1/7B= 5/2.7 + 4/7.11 + 3/11.14 + 1/14.15 + 13/15.28
1/7B= 1/2 - 1/7 + 1/7 - 1/11 + 1/11 - 1/14 + 1/14 - 1/15 + 1/15 - 1/28
1/7B= 1/2 - 1/28= 13/28
B= 13/28 : 1/7= 13/28.7= 13/4
Vậy B= 13/4
E=\(\frac{5}{1.2}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
E.\(\frac{1}{7}\)=\(\frac{5}{1.2.7}+\frac{4}{1.11.7}+\frac{3}{11.2.7}+\frac{1}{2.15.7}+\frac{13}{15.4.7}\)
E.\(\frac{1}{7}\)=\(\frac{5}{7.2}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
E.\(\frac{1}{7}\)=\(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
E.\(\frac{1}{7}\)=\(\frac{1}{2}-\frac{1}{28}\)
E.\(\frac{1}{7}=\frac{13}{28}\)
E=\(\frac{13}{28}:\frac{1}{7}=\frac{13}{4}\)
\(E=\frac{5}{1.2}+\frac{1}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(E=\frac{5}{2}+\frac{1}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\)
\(E=\frac{5}{2}+\left(\frac{1}{11}+\frac{3}{22}\right)+\left(\frac{1}{30}+\frac{13}{60}\right)\)
\(E=\frac{5}{2}+\left(\frac{2}{22}+\frac{3}{22}\right)+\left(\frac{2}{60}+\frac{13}{60}\right)\)
\(E=\frac{5}{2}+\frac{5}{22}+\frac{15}{60}\)
\(E=\frac{55}{22}+\frac{5}{22}+\frac{1}{4}\)
\(E=\frac{60}{22}+\frac{1}{4}\)
\(E=\frac{30}{11}+\frac{1}{4}\)
\(E=\frac{120}{44}+\frac{11}{44}\)\(=\frac{131}{44}\)
k mình nha chúc bạn học giỏi
\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)
\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)
\(=\frac{2^9.3^9.-17}{1}\)
Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)
\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3.2^{18}}{2^{35}.3^2}\)
\(=\frac{1}{2^{17}.3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\frac{1}{10.11}+7.\frac{1}{11.12}+7.\frac{1}{12.13}+...+7.\frac{1}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=\frac{6}{70}\)
\(=\frac{3}{35}\)
a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}=\frac{31}{32}\)
b) \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)\
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\)\(\frac{3}{11\cdot14}+...+\)\(\frac{3}{602\cdot605}\)
\(=\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{602\cdot605}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{602}\)\(-\frac{1}{605}\)
\(=\frac{1}{5}-\frac{1}{605}\)
\(=\frac{121}{605}-\frac{1}{605}\)
\(=\frac{120}{605}=\frac{24}{121}\)
Bài này dùng công thức nhé
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{602.605}\)
\(=\)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{602}-\frac{1}{605}\)
\(=\)\(\frac{1}{5}-\frac{1}{605}\)
\(=\)\(\frac{24}{121}\)
Chúc bạn học tốt ~