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Ta có: \(S=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
\(=\dfrac{3}{2^2}\cdot\dfrac{2^3}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99}{10^2}\)
\(=\dfrac{11}{20}\)
bạn có thể giải thích rõ tại sao S=\(\dfrac{11}{20}\) đc ko
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)
10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)
\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)
\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)
\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)
\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)
\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)
Giải:
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}.\)
\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{29.31}{30^2}.\)
\(A=\dfrac{1.2.3.....29}{2.3.4.....30}.\dfrac{2.3.4.....31}{2.3.4.....30}.\)
\(A=\dfrac{1}{30}.31=\dfrac{30}{31}.\)
Vậy \(A=\dfrac{30}{31}.\)
a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)
b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)
c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)
c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)
\(a.\)
\(\dfrac{3}{16}:\dfrac{?}{8}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{16}\cdot\dfrac{8}{?}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{2?}=\dfrac{3}{4}\)
\(\Leftrightarrow?=2\)
\(b.\)
\(\dfrac{1}{25}:-\dfrac{3}{?}=-\dfrac{1}{15}\)
\(\Leftrightarrow\dfrac{1}{25}\cdot\dfrac{-?}{3}=-\dfrac{1}{15}\)
\(\Leftrightarrow\dfrac{-?}{75}=-\dfrac{1}{15}\)
\(\Leftrightarrow?=\dfrac{75}{15}=5\)
\(c.\)
\(\dfrac{?}{12}:-\dfrac{4}{9}=-\dfrac{3}{16}\)
\(\Leftrightarrow\dfrac{?}{12}\cdot\dfrac{-9}{4}=-\dfrac{3}{16}\)
\(\Leftrightarrow\dfrac{-3?}{16}=-\dfrac{3}{16}\)
\(\Leftrightarrow?=1\)
Mk gọi ? = x nha
a) \(\dfrac{3}{16}:\dfrac{x}{8}=\dfrac{3}{4}\)
\(\dfrac{x}{8}=\dfrac{3}{16}:\dfrac{3}{4}\)
\(\dfrac{x}{8}=\dfrac{1}{4}\)
⇒\(x=\dfrac{1.8}{4}=2\)
b) \(\dfrac{1}{25}:\dfrac{-3}{x}=\dfrac{-1}{15}\)
\(\dfrac{-3}{x}=\dfrac{1}{25}:\dfrac{-1}{15}\)
\(\dfrac{-3}{x}=\dfrac{-3}{5}\)
⇒x=5
c) \(\dfrac{x}{12}:\dfrac{-4}{9}=\dfrac{-3}{16}\)
\(\dfrac{x}{12}=\dfrac{-3}{16}.\dfrac{-4}{9}\)
\(\dfrac{x}{12}=\dfrac{1}{12}\)
⇒x=1
a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)
\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)
\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)
b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)
=459/940
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.\dfrac{4.6}{5^2}...\dfrac{8.10}{9^2}.\dfrac{9.11}{10^2}\)
\(=\dfrac{1.2.3.4...8.9}{2.3.4.5...10}.\dfrac{3.4.5.6...11}{2.3.4.5...10}\)
\(=\dfrac{1}{10}.\dfrac{11}{2}\)
\(=\dfrac{11}{20}\)
Ta có:
\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}....\dfrac{80}{81}.\dfrac{99}{100}\\ =\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.\dfrac{4.6}{5^2}...\dfrac{8.10}{9^2}.\dfrac{9.11}{10^2}\\ =\dfrac{11}{2.10}=\dfrac{11}{20}\)