Các câu hỏi tương tự
BD
18 tháng 3 2023 lúc 19:38
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
NM
11 tháng 5 2022 lúc 14:10
tính nhanh \(S=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)
NM
11 tháng 5 2022 lúc 14:21
tính nhanh \(S=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)
NM
11 tháng 5 2022 lúc 14:13
tính nhanh \(S=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)
\(\dfrac{3}{2}X-0,2=\dfrac{3}{5}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{11}{8}-\dfrac{3}{8}.x=\dfrac{1}{8}\)
giúp với
\((\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{9}+\dfrac{1}{10})\times x=\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}\)
Bài 1:Chứng tỏ rằng:Bdfrac{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}1Bài 2:Chứng tỏ rằng:Edfrac{3}{4}+dfrac{8}{9}+dfrac{15}{16}+...+dfrac{2499}{2500}1Bài 3:Chứng tỏ rằng:1dfrac{2011}{2020^2+1}+dfrac{2021}{2020^2+2}+dfrac{2021}{2020^3+3}+...+dfrac{2021}{2020^3+2020} 2
Đọc tiếp
Bài 1:Chứng tỏ rằng:B=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)\(\dfrac{1}{8^2}\)<1
Bài 2:Chứng tỏ rằng:E=\(\dfrac{3}{4}\)+\(\dfrac{8}{9}\)+\(\dfrac{15}{16}\)+...+\(\dfrac{2499}{2500}\)<1
Bài 3:Chứng tỏ rằng:1<\(\dfrac{2011}{2020^2+1}\)+\(\dfrac{2021}{2020^2+2}\)+\(\dfrac{2021}{2020^3+3}\)+...+\(\dfrac{2021}{2020^3+2020}\)< 2
b)\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{9}+\dfrac{1}{10}\right)x=\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}\)
Tính hợp lý
\(A= (\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\) B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}})\)