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\(\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{8}\right)\times\left(1+\frac{1}{15}\right)\times...\times\left(1+\frac{1}{9999}\right)\)
\(=\frac{2^2}{1\cdot3}\times\frac{3^2}{2\cdot4}\times\frac{4^2}{3\cdot5}\times...\times\frac{100^2}{99\cdot101}\)
\(=\frac{2\cdot3\cdot4\cdot...\cdot100}{1\cdot2\cdot3\cdot...\cdot99}\times\frac{2\cdot3\cdot4\cdot...\cdot100}{3\cdot4\cdot5\cdot...\cdot101}\)
\(=\frac{100}{1}\times\frac{2}{101}=\frac{200}{101}.\)
( 4.2/5 +3/5) + (2.3/4 +1/4) +(5.2/3 +1/3) =( 4 + 2/5 + 3/5) +( 2 +3/4 + 1/4 ) + (5 + 2/3 + 1/3) =( 4+1 ) + (2 +1) + ( 5 + 1) = 14
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
\(A=\frac{2}{3}+\frac{1}{3}+\frac{1}{15}+\frac{1}{75}+\frac{1}{375}+\frac{1}{1875}\)
\(A=\left(\frac{2}{3}+\frac{1}{3}\right)+\frac{1}{15}+\frac{1}{75}+\frac{1}{375}+\frac{1}{1875}\)
\(A=1+\left(\frac{5}{75}+\frac{1}{75}\right)+\frac{1}{375}+\frac{1}{1875}\)
\(A=1+\frac{6}{75}+\frac{1}{375}+\frac{1}{1875}\)
\(A=1+\left(\frac{150}{1875}+\frac{1}{1875}\right)+\frac{1}{375}\)
\(A=1+\frac{151}{1875}+\frac{1}{375}\)
\(A=1+\left(\frac{151}{1875}+\frac{5}{1875}\right)\)
\(A=1+\frac{156}{1875}\)
\(A=\frac{1875}{1875}+\frac{156}{1875}\)
\(A=\frac{677}{625}\)
\(A=\frac{2}{3}+\frac{1}{3}+\frac{1}{15}+\frac{1}{75}+\frac{1}{375}+\frac{1}{1875}\)
\(\Leftrightarrow A=\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{1}{15}+\frac{1}{75}\right)+\left(\frac{1}{375}+\frac{1}{1875}\right)\)
\(\Leftrightarrow A=1+\left(\frac{5}{75}+\frac{1}{75}\right)+\left(\frac{5}{1875}+\frac{1}{1875}\right)\)
\(\Leftrightarrow A=1+\frac{6}{75}+\frac{6}{1875}\)
\(\Leftrightarrow A=1+\frac{2}{25}+\frac{2}{625}\)
\(\Leftrightarrow A=1+\left(\frac{2}{25}+\frac{2}{625}\right)\)
\(\Leftrightarrow A=1+\left(\frac{50}{625}+\frac{2}{625}\right)\)
\(\Leftrightarrow A=1+\frac{52}{625}\)
\(\Leftrightarrow A=\frac{677}{625}\)
\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}\cdot\frac{4}{7}\cdot\frac{7}{10}\cdot\cdot\cdot\frac{97}{100}\)
\(=\frac{1.4.7.10...97}{4.7.10.13...100}\)
\(=\frac{1}{100}\)
\(1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot1\frac{1}{24}\cdot...\cdot1\frac{1}{99}\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot\frac{25}{24}\cdot...\cdot\frac{100}{99}\)
\(=\frac{2.2\cdot3.3\cdot4.4\cdot5.5\cdot...\cdot10.10}{1.3\cdot2.4\cdot3.5\cdot4.6\cdot...\cdot9.11}\)
\(=\frac{2.10}{1.11}=\frac{20}{11}\)
"." = nhân