Chứng minh : \(\dfrac{99}{100}\) > \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)
Tìm x, biết:
a).\(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{10.110}\right).x=\dfrac{1}{1.11}+\dfrac{1}{2.12}+...+\dfrac{1}{100.110}\)
b).\(x-\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-...-\dfrac{20}{53.55}=\dfrac{3}{11}\)
c).\(\dfrac{x-1}{99}+\dfrac{x-2}{98}+\dfrac{x-5}{95}=3+\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)
Mấy bạn tính nhanh, hợp lí, giải ra từng bước dùm mik nha
Thanks m.n
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)
Chứng minh rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
Chứng tỏ rằng:
a) \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{2}\)
b) \(S=\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{80}>\dfrac{7}{12}\)
c) \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
d) \(\dfrac{49}{100}< S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}< 1\)
Các bạn giải ra từng bước dùm mik nha
Thanks m.n
1.Tính nhanh:
A= \(\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}\)
2. Cho: B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\) .Hãy chứng tỏ rằng B > 1.
3. Rút gọn:
a) C= \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{20}\right)\)
b) D= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}\)
4. So sánh: E=\(\dfrac{20^{10}+1}{20^{10}-1}\) và F =\(\dfrac{20^{10}-1}{20^{10}-3}\)
5. Tính giá trị của biểu thức:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
Tính giá trị của biểu thức: Q =\((\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999})\cdot(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6})\)
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(X=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{99^2}{98.100}\)
\(K=\dfrac{1}{3}.\dfrac{1}{15}.\dfrac{1}{35}...\dfrac{1}{9999}\)
Bài 1 : Thực hiện phép tính ( tính hợp lý nếu có thể )
a ) \(\dfrac{1}{12}+\dfrac{3}{4}\)
b ) \(\dfrac{-4}{7}.1\dfrac{1}{2}\)
c )\(\dfrac{7}{9}+\left(\dfrac{2}{3}+\dfrac{-7}{9}\right)\)
d )\(\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}\)
e )\(\dfrac{-7}{25}.\dfrac{11}{13}+\dfrac{-7}{25}.\dfrac{2}{13}\)
g )\(2\dfrac{2}{5}.0,25-\left(\dfrac{11}{20}+75\%\right):\dfrac{13}{5}\)