C=\(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\)
E=\(\dfrac{6}{1.3.7}+\dfrac{6}{6.7.9}+...+\dfrac{6}{13.15.19}\)
F=\(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}}{\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}}\)
G=\(10-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)
\(C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\)
\(C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{21}\)
\(C=\dfrac{2}{7}\)
Tìm x;
a,\(\left(1-\dfrac{1}{3}\right)\)*\(\left(1-\dfrac{1}{6}\right)\)*\(\left(1-\dfrac{1}{10}\right)\)*...*\(\left(1-\dfrac{1}{780}\right)\)*x=1
Ta có:
\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right).x=1\)
\(\Leftrightarrow\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}.x=1\)
\(\Leftrightarrow\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}.....\dfrac{1558}{1560}.x=1\)
\(\Leftrightarrow\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}.......\dfrac{38.41}{39.40}.x=1\)
\(\Leftrightarrow\dfrac{1.4.2.5.3.6.....38.41}{2.3.3.4.4.5.....39.40}.x=1\)
\(\Leftrightarrow\dfrac{\left(1.2.3.4....38\right)\left(4.5.6.7....41\right)}{\left(2.3.4.....39\right)\left(3.4.5....40\right)}.x=1\)
\(\Leftrightarrow\dfrac{1}{39}.\dfrac{41}{3}.x=1\)
\(\Leftrightarrow\dfrac{41}{117}.x=1\)
\(\Leftrightarrow x=\dfrac{117}{41}\)
Vậy ...
\(\dfrac{1\cdot2\cdot3+2\cdot4+6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\)
\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)
Bạn ghi đề sai rồi, mình sửa lại đề ở phần (*) rồi nhé! 
Ta có: \(\dfrac{1.2.3+2.4.6+4.8.12}{1.3.5+2.6.10+4.12.20}\) (*)
= \(\dfrac{1.2.3\left(1+2^3+4^3\right)}{1.3.5\left(1+2^3+4^3\right)}\) = \(\dfrac{1.2.3}{1.3.5}\) = \(\dfrac{2}{5}\)
\(1\dfrac{1}{3}\cdot1\dfrac{1}{8}\cdot1\dfrac{1}{15}\cdot....\cdot1\dfrac{1}{99}\)
\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}......1\dfrac{1}{99}\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)
\(=\dfrac{2.2.3.3.4.4.....10.10}{1.3.2.4.3.5.....9.11}\) ( Bước này bạn bỏ đi cũng được )
\(=\dfrac{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}\)
\(=\dfrac{\left(2.3.4.....9\right).10.2.\left(3.4.5.....10\right)}{1.\left(2.3.4.....9\right).\left(3.4.5.....10\right).11}\)
\(=\dfrac{10.2}{1.11}=\dfrac{20}{11}=1\dfrac{9}{11}\)
\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.....1\dfrac{1}{99}\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)
\(=\dfrac{2.2.3.3.4.4.....10.10}{1.3.2.4.3.5....9.11}\)
\(=\dfrac{2.3.4....10}{1.2.3....9}.\dfrac{2.3.4...10}{3.4.5....11}\)
\(=10.\dfrac{2}{11}=\dfrac{20}{11}\)
\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.....1\dfrac{1}{99}\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)
\(=\dfrac{2.3.4.....10}{1.2.3.....9}.\dfrac{2.3.4.....10}{3.4.5.....11}\)
\(=10.\dfrac{2}{11}\)
\(=\dfrac{20}{11}\)
\(\dfrac{1}{3\cdot63}\)+\(\dfrac{1}{4\cdot64}\)+...+\(\dfrac{1}{1802\cdot1862}\)
(xoxox x3):x -20002
Tính nhanh:
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).....\left(\dfrac{1}{99}-1\right)\)
Ta có : \(\dfrac{-1}{2}\).\(\dfrac{-2}{3}\)\(\dfrac{-3}{4}\). ... . \(\dfrac{-98}{99}\)
=>\(\dfrac{\left(-1\right).\left(-2\right).\left(-3\right).....\left(-98\right)}{2.3.4.....99}\)
=> \(\dfrac{1}{99}\)
Tick mình nha bạn hiền.
(1/2-1) (1/3-1) (1/4-1) ... (1/99-1)
=(-1/2) (-2/3) (-3/4) ... (-98/99)
=(-1/99)
Vậy (1/2-1) (1/3-1) (1/4-1) ... (1/99-1)=(-1/99)
Đặt \(A\) \(=\) \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{99}-1\right)\)
\(A=\left(-\dfrac{1}{2}\right)\left(-\dfrac{2}{3}\right)\left(-\dfrac{3}{4}\right)...\left(-\dfrac{98}{99}\right)\)
\(A=\dfrac{1.2.3.....98}{2.3.4.....99}\)
\(A=\dfrac{1}{99}\)
\(\dfrac{1}{3\cdot1803}\)+\(\dfrac{1}{4\cdot1804}\)+...+\(\dfrac{1}{62\cdot1862}\)
