Violympic toán 7
Các câu hỏi tương tự
Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Chứng minh rằng:
\(\dfrac{1}{6}\) < \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\)
Cho A=\(\dfrac{2}{1}.\dfrac{4}{3}.\dfrac{6}{5}.\dfrac{8}{7}.\dfrac{10}{9}...\dfrac{100}{99}\). Chứng minh rằng 12<A<13
cho A =\(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)
Chứng minh rằng A<\(\dfrac{1}{3}\)
13 tháng 10 2019 lúc 21:01
Chứng minh rằng :
\(\dfrac{1}{7^2} - \dfrac{1}{7^4} + ... + \dfrac{1}{7^{4n-2}} - \dfrac{1}{7^{4n}} + ... + \dfrac{1}{7^{98}}-\dfrac{1}{7^{100}} < \dfrac{1}{50}\)
CMR: \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
1. Chứng minh rằng:
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100}< 1\)
2. Chứng minh rằng:
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100-1}{100!}< 2\)
2 tháng 7 2021 lúc 21:10
Tính giá trị biểu thức A , biết rằng A M : N Mà M dfrac{dfrac{1}{99}+dfrac{2}{98}+dfrac{3}{97}+dfrac{4}{96}+...+dfrac{97}{3}+dfrac{98}{2}+dfrac{99}{1}}{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+dfrac{1}{5}+dfrac{1}{6}+...+dfrac{1}{100}} N dfrac{92-dfrac{1}{9}-dfrac{2}{10}-dfrac{3}{11}-...-dfrac{90}{98}-dfrac{91}{99}-dfrac{92}{100}}{dfrac{1}{45}+dfrac{1}{50}+dfrac{1}{55}+...+dfrac{1}{495}+dfrac{1}{500}}
Đọc tiếp
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)