a)

\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)

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

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

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

\(\left(1-\dfrac{28}{10}\right)\left(1-\dfrac{52}{22}\right)\left(1-\dfrac{80}{36}\right)...\left(1-\dfrac{21808}{10900}\right)\)

\(=\left(1-\dfrac{2.10+8}{10}\right)\left(1-\dfrac{2.22+8}{22}\right)\left(1-\dfrac{2.36+8}{36}\right)...\left(1-\dfrac{2.10900+8}{10900}\right)\)

\(=\left(1-2-\dfrac{8}{10}\right)\left(1-2-\dfrac{8}{22}\right)\left(1-2-\dfrac{8}{36}\right)...\left(1-2-\dfrac{8}{10900}\right)\)

\(=\left(-1-\dfrac{8}{1.10}\right)\left(-1-\dfrac{8}{2.11}\right)\left(-1-\dfrac{8}{3.12}\right)...\left(-1-\dfrac{8}{100.109}\right)\)

\(=\left(\dfrac{-18}{1.10}\right)\left(\dfrac{-30}{2.11}\right)\left(\dfrac{-44}{3.12}\right)...\left(\dfrac{-10908}{100.109}\right)\)

\(=\left(\dfrac{-2.9}{1.10}\right)\left(\dfrac{-3.10}{2.11}\right)\left(\dfrac{-4.11}{3.12}\right)...\left(\dfrac{-101.108}{100.109}\right)\)

\(=\dfrac{\left(-2\right)\left(-3\right)\left(-4\right)...\left(-101\right)}{1.2.3...100}.\dfrac{9.10.11...108}{10.11.12...109}\) (1)

\(=\dfrac{101}{1}.\dfrac{9}{109}=\dfrac{909}{109}\)

Do ở (1) có \(-2-\left(-101\right)+1=100\) nhân tử (số nhân tử là số chẵn) mang dấu âm nên kết quả sẽ mang dấu dương

bấn nhần ảnh xin lỗi

\(A=\left(1-\dfrac{1}{15}\right).\left(1-\dfrac{1}{21}\right).\left(1-\dfrac{1}{28}\right)....\left(1-\dfrac{1}{210}\right)\)

\(A=\dfrac{14}{15}.\dfrac{20}{21}.\dfrac{27}{28}....\dfrac{209}{210}\)

\(A=\dfrac{14.2}{15.2}.\dfrac{20.2}{21.2}.\dfrac{27.2}{28.2}....\dfrac{209.2}{210.2}\)

\(A=\dfrac{4.7}{5.6}.\dfrac{5.8}{6.7}.\dfrac{6.9}{7.8}....\dfrac{19.22}{20.21}\)

\(A=\dfrac{4.5.6....19}{5.6.7....20}.\dfrac{7.8.9....22}{6.7.8....21}\)

\(A=\dfrac{4}{20}.\dfrac{22}{6}\)

\(A=\dfrac{11}{15}\)

a: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{2}{17}\)

b: \(=-8\cdot\dfrac{1}{4}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-2:\dfrac{27-14}{12}=\dfrac{-2\cdot12}{13}=-\dfrac{24}{13}\)

c: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)\)

=-520/7

mk chỉ giải đc câu d thôi nha ; bn thông cảm

d) \(D=\dfrac{2006.2005-1}{2004.2006+2005}=\dfrac{2006.2005-1}{2004.2006+2006-1}\)

\(D=\dfrac{2006.2005-1}{2006\left(2004+1\right)-1}=\dfrac{2006.2005-1}{2006.2005-1}=1\)

e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)

=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)

=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)

=\(-\dfrac{516}{7}\)

a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)

=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)

=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)

=\(\dfrac{119}{240}\)

b)\(\dfrac{3}{2}-\dfrac{5}{6}:\left(\dfrac{1}{2}\right)^2+\sqrt{4}\)

=\(\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+2\)

=\(\dfrac{3}{2}-\dfrac{10}{3}+2\)

=\(-\dfrac{11}{6}\) +2

=\(\dfrac{1}{6}\)

Ta có: \(A=\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{15}-1\right)\left(\dfrac{1}{21}-1\right)\left(\dfrac{1}{28}-1\right)\left(\dfrac{1}{36}-1\right)\)

\(=\dfrac{-2}{3}.\dfrac{-5}{6}.\dfrac{-9}{10}.\dfrac{-14}{15}.\dfrac{-20}{21}.\dfrac{-27}{28}.\dfrac{-35}{36}\)

\(=\dfrac{-2.\left(-5\right).3.\left(-3\right).2.\left(-7\right).\left(-4\right).5.\left(-3\right).9.5.\left(-7\right)}{3.2.3.2.5.3.5.3.7.4.7.4.9}\)

\(=\dfrac{-5}{3.4}=\dfrac{-5}{12}\)

Vậy \(A=\dfrac{-5}{12}.\)

\(C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\)

\(2C=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)

\(2C=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{2015}}\)

\(2C-C=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2015}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)\)

\(C=2-\dfrac{1}{2^{2016}}\)

trẻ trâu lửa chùa

hahahaha

ăn lồn cái địt mẹ mày lửa trùa ĐẦU LỒN nhá

pro ĐẦU LỒN mày nhá

Mk ko biết nhưng mk chúc bn sớm tìm đc câu trả lời

b)

\(\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{2012}-1\right)\\ =\dfrac{-9}{10}\cdot\dfrac{-10}{11}\cdot\dfrac{-11}{12}\cdot...\cdot\dfrac{-2011}{2012}\\ =\left(-1\right)\cdot\dfrac{9}{10}\cdot\left(-1\right)\cdot\dfrac{10}{11}\cdot\left(-1\right)\cdot\dfrac{11}{12}\cdot...\cdot\left(-1\right)\cdot\dfrac{2011}{2012}\\ =\left[\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)\right]\cdot\left(\dfrac{9}{10}\cdot\dfrac{10}{11}\cdot\dfrac{11}{12}\cdot...\cdot\dfrac{2011}{2012}\right)\\ =\left[\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)\right]\cdot\dfrac{9}{2012}\)

(Có tất cả 2003 thừa số -1)

\(=\left(-1\right)\cdot\dfrac{9}{2012}=\dfrac{-9}{2012}\)