tính S=\(\dfrac{3}{4}\)*\(\dfrac{8}{9}\)*\(\dfrac{15}{16}\)*........*\(\dfrac{99}{100}\)
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Khách
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HH
\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}._{......}.\dfrac{80}{81}.\dfrac{99}{100}\)
2
LG
22 tháng 5 2017
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.\dfrac{4.6}{5^2}...\dfrac{8.10}{9^2}.\dfrac{9.11}{10^2}\)
\(=\dfrac{1.2.3.4...8.9}{2.3.4.5...10}.\dfrac{3.4.5.6...11}{2.3.4.5...10}\)
\(=\dfrac{1}{10}.\dfrac{11}{2}\)
\(=\dfrac{11}{20}\)
22 tháng 5 2017
Ta có:
\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}....\dfrac{80}{81}.\dfrac{99}{100}\\ =\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.\dfrac{4.6}{5^2}...\dfrac{8.10}{9^2}.\dfrac{9.11}{10^2}\\ =\dfrac{11}{2.10}=\dfrac{11}{20}\)
R
30 tháng 3 2018
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
6 tháng 3 2018
\(\left(a\right):P=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{99}{100}\)
Nhận xét
thừa số tổng quát là \(\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\) với n =1 đến 10
\(P=\dfrac{1.3.2.4.3.5...9.11}{2^2.3^2...9^2.10^2}=\dfrac{\left(1.2.3...9\right)\left(3.4.5....11\right)}{\left(2.3.4....10\right)\left(2.3.4....10\right)}\)
\(P=\dfrac{1.2.3..9}{2.3.4..9.10}.\dfrac{3.4.5...10.11}{2.3.4....10}=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)
TT
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 11 2018
Đặt \(A=\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{9999}{10000}=1-\dfrac{1}{4}+1-\dfrac{1}{9}+...+1-\dfrac{1}{10000}\)
\(=99-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)=99-B\)
Do \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>0\Rightarrow99-B< 99\Rightarrow A< 99\)
Do \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}\)
\(\Rightarrow A=99-B>99-\left(1-\dfrac{1}{100}\right)=98+\dfrac{1}{100}>98\)
Vậy \(98< \dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{9999}{10000}< 99\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 10 2023
\(\dfrac{1}{2}-\dfrac{3}{8}=\dfrac{4}{2\times4}-\dfrac{3}{8}=\dfrac{4}{8}-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\dfrac{4}{3}-\dfrac{8}{15}=\dfrac{4\times5}{3\times5}-\dfrac{8}{15}=\dfrac{20}{15}-\dfrac{8}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\dfrac{5}{6}-\dfrac{7}{12}=\dfrac{5\times2}{6\times2}-\dfrac{7}{12}=\dfrac{10}{12}-\dfrac{7}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)
\(\dfrac{11}{4}-\dfrac{9}{8}=\dfrac{11\times2}{4\times2}-\dfrac{9}{8}=\dfrac{22}{8}-\dfrac{9}{8}=\dfrac{13}{8}\)
\(\dfrac{17}{16}-\dfrac{3}{4}=\dfrac{17}{16}-\dfrac{3\times4}{4\times4}=\dfrac{17}{16}-\dfrac{12}{16}=\dfrac{5}{16}\)
\(\dfrac{31}{36}-\dfrac{5}{6}=\dfrac{31}{36}-\dfrac{5\times6}{6\times6}=\dfrac{31}{36}-\dfrac{30}{36}=\dfrac{1}{36}\)
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