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\(A=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.......\frac{2907}{2900}\)
\(A=\frac{2.8}{1.9}.\frac{3.9}{2.10}.\frac{4.10}{3.11}......\frac{51.57}{50.58}\)
\(A=\frac{2.3.4.....51}{1.2.3...50}.\frac{8.9.10....57}{9.10.11...58}\)
\(A=51.\frac{8}{58}=\frac{204}{29}\)
Bạn Nguyễn Tuấn Minh làm đúng rùi đó !!! Chuẩn ý kiến mk...^.^
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2}{3}...\frac{2017}{2018}\)
\(=\frac{1.2...2017}{2.3...2018}\)
\(=\frac{1}{2018}\)
b) \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{190}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{189}{190}\)
\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{378}{380}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{7.4}{5.6}...\frac{18.21}{19.20}\)
\(=\frac{\left(1.2.3...18\right).\left(4.5.6...21\right)}{\left(2.3.4...19\right).\left(3.4.5...20\right)}\)
\(=\frac{1.21}{19.3}\)
\(=\frac{21}{57}\)
c) \(\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)\left(1+\frac{7}{48}\right)...\left(1+\frac{7}{2009}\right)\)
\(=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.\frac{56}{48}...\frac{2016}{2009}\)
mk ko bít làm câu c ! xin lỗi bn nha! bn tự nghĩ cách làm câu c giúp mk nhé!
Cô giải như sau Minh nhé :)
\(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{2900}\right)=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}...\frac{2907}{2900}\)
\(=\frac{8.2}{9.1}.\frac{9.3}{10.2}.\frac{10.4}{11.3}....\frac{57.51}{58.50}=\frac{\left(8.9.10....57\right)\left(2.3.4...51\right)}{\left(9.10.11...58\right)\left(2.3.4....50\right)}=\frac{8.51}{58}=\frac{204}{29}\)
( 1 + 7/9 ) x ( 1 + 7/20 ) x ( 1 + 7/33 ) x...x ( 1 + 7/2900)
= (8x2)/(9x1) x (9x3)/(10x2) x (10x4)/(11x3) x...x (57x51)(58x50)
=(8x2x9x3x10x4x...x57x51) / (9x1x10x2x11x3x...x58x50) Sau khi giản ước ta được :
= (8x51) / (1x58) = 204/29
567
567
100% tick nha