A = \(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}=\frac{1\cdot3}{2.2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99\cdot101}{100\cdot100}=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)

B = ( 1- 1/4 )( 1-1/9) ...( 1-1/10000 ) = 3/4 . 8/9 .....9999/100000 ( tương tự A )

a=5051/100 co ma

xin loi ban lam dung roi minh nham

mình làm được nhưng đánh lâu lắm

biết mỗi câu a

ukm thế cx dc

cau a bang 101/200, k cho minh nha'

\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)

\(B=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\)

\(B=\frac{3.8.15...9999}{2^2.3^2.4^2...100^2}\)

\(B=\frac{1.3.2.4.3.5...99.101}{2.2.3.3.4.4...100.100}\)

\(B=\frac{\left(1.2.3...99\right).\left(3.4.5...101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}\)

\(B=\frac{1.101}{100.2}\)

\(B=\frac{101}{200}\)

          \(C=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right).\left(1+\frac{1}{100}\right)\)

           \(C=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}.\frac{101}{100}\)

            \(C=\frac{3.4.5...100.101}{2.3.4...99.100}\)

           \(C=\frac{101}{2}\)

 Dấu . là dâú x nha

\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(3A=3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)

\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(3A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)

\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(3A=\frac{1}{2}-\frac{1}{20}\)

\(A=\left(\frac{1}{2}-\frac{1}{20}\right)\div3=\frac{9}{20}\div3=\frac{9}{20.3}=\frac{3}{20}\)

Vậy ................

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot....\cdot\frac{9999}{10000}\)

\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{99.101}{100.100}\)

\(B=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right).\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right).\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)

\(B=\frac{1\cdot2\cdot3\cdot..\cdot99}{2\cdot3\cdot4\cdot..\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)

\(B=\frac{1}{100}\cdot\frac{101}{2}=\frac{101}{200}\)

vậy......

A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20

A=1/3.(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17+3/17.20)

A=1/3.(1/2-1/20)

=3/20

B=1.3/2.2+2.4/3.3+3.5/4.4+...+99.101/100.100

B=(1.2.3...99).(3.4.5...101)/(2.3.4...100).(2.3.4...100)

B=\(\frac{1.2....99}{2.3...100}\).\(\frac{3.4...101}{2.3...100}\)

B=1/100.101/2=101/200

\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{99\cdot101}{100\cdot100}\\ =\dfrac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}\\ =\dfrac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}\\ =\dfrac{1\cdot101}{100\cdot2}\\ =\dfrac{101}{200}\)

\(C=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}=\dfrac{200}{101}\)

\(B=\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot...\cdot\left(1-\dfrac{1}{1326}\right)\\ =\dfrac{20}{21}\cdot\dfrac{27}{28}\cdot\dfrac{35}{36}\cdot...\cdot\dfrac{1325}{1326}\\=\dfrac{40}{42}\cdot\dfrac{54}{56}\cdot\dfrac{70}{72}\cdot...\cdot\dfrac{2650}{2652}\\ =\dfrac{5\cdot8}{6\cdot7}\cdot\dfrac{6\cdot9}{7\cdot8}\cdot\dfrac{7\cdot10}{8\cdot9}\cdot...\cdot\dfrac{50\cdot53}{51\cdot52}\\ =\dfrac{5\cdot8\cdot6\cdot9\cdot7\cdot10\cdot...\cdot50\cdot53}{6\cdot7\cdot7\cdot8\cdot8\cdot9\cdot...\cdot51\cdot52}\\ =\dfrac{\left(5\cdot6\cdot7\cdot...\cdot50\right)\cdot\left(8\cdot9\cdot10\cdot...\cdot53\right)}{\left(6\cdot7\cdot8\cdot...\cdot51\right)\cdot\left(7\cdot8\cdot9\cdot...\cdot52\right)}=\dfrac{5\cdot53}{51\cdot7}=\dfrac{265}{357} \)

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}......\frac{99.101}{100.100}\)

\(=\frac{1.2.3...99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)

\(=\frac{1}{100}.\frac{101}{2}\)

\(=\frac{101}{200}\)