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.
tìm a,b biết :
\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
tính a, \(\dfrac{5.4^{15}.9^9-4.30^{20}8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
b, 1\(\dfrac{1}{30}\):\(\left(24\dfrac{1}{6}-24\dfrac{1}{5}\right)-\dfrac{1\dfrac{1}{2}-\dfrac{3}{4}}{4x-\dfrac{1}{2}}=\left(-1\dfrac{1}{15}\right):\left(8\dfrac{1}{5}-8\dfrac{1}{3}\right)\)
Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)
a, Cho b là số tự nhiên, b>1. Chứng minh rằng: \(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b^2}< \dfrac{1}{b-1}-\dfrac{1}{b}\)
b, Áp dụng phần a: Cho S\(=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\). Chứng minh rằng: \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
Tính tổng sau:
a) \(\dfrac{1}{9}+3,25+5\dfrac{3}{16}+4\dfrac{1}{3}+2,8+0,5\)
b) \(2\dfrac{1}{3}+0,45+4,25+\dfrac{1}{81}+6\dfrac{8}{27}\)
c) \(1,25+2\dfrac{1}{4}+4\dfrac{2}{5}+0,3+2,14+4\dfrac{1}{8}\)
Bài1. (4điểm) Thực hiện phép tính:
a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
b) \(B=\dfrac{-1}{2}+\dfrac{-1}{6}+\dfrac{-1}{12}+\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
Tìm \(a,b\in Z\) biết \(\dfrac{a}{9}-\dfrac{3}{b}=\dfrac{1}{18}\)
CMR : \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Với \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\)