A= \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\)
= 1-\(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{6}\)
= 1 - \(\dfrac{1}{6}\)= \(\dfrac{5}{6}\)
mk chỉ bt làm câu 1 thôi ak
mong bn thông cảm
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)
Giup với mai KT ròi.
Thanks nha.
\(\text{Tìm x, biết:}\)
\(a\)) \(x-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-\dfrac{2}{9.11}-\dfrac{2}{11.13}-\dfrac{2}{13.15}=\dfrac{2}{5}\)
\(b\)) \(\dfrac{1}{2.3}.x+\dfrac{1}{3.4}.x+\dfrac{1}{4.5}.x+...+\dfrac{1}{49.50}.x=1\)
\(c\)) \(x-\dfrac{20}{11.3}-\dfrac{20}{13.15}-...-\dfrac{53}{55}=\dfrac{3}{11}\)
\(d\)) \(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=\dfrac{-37}{45}\)
\(e\)) \(\left(\dfrac{11}{12}.\dfrac{11}{2.23}.\dfrac{11}{23.34}...\dfrac{11}{89.100}\right).x=\dfrac{5}{3}\)
\(f\)) \(\left(\dfrac{2}{11.13}.\dfrac{2}{13.15}.\dfrac{2}{15.17}...\dfrac{2}{19.21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
Cho A=\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{59.60}\) Chứng tỏ A > \(\dfrac{7}{12}\)
Các bạn giúp với :<
Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
Tính: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\)
Tính A:
A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
HELP ME!
Tính tổng
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
