1) Cho \(A=\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}.CMR:A< \frac{1}{9!}\)
2) \(CMR:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ai giúp mk sẽ đc thưởng 3 tick , phải ghi chép đầy đủ nha
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Bạn tham khảo nhé
\(a)\)Đặt \(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}< 1\) ( đpcm )
Vậy \(A< 1\)
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}
a)Cho S = \(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2012!}.\) Chứng minh rằng S< 2
b)Chứng minh rằng :\(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+\frac{99}{100!}< \frac{1}{9!}\)
Ai làm nhanh mk l*** cho nhé !
sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)
Bài 1;Cho S = \(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.....................+\frac{1}{2012!}\)CMR: S <2
Bài 2:CMR \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...........+\frac{99}{100!}<\frac{1}{9!}\)
Bài 3: Cho E= \(1+\frac{1}{2}+\frac{1}{3}+...........+\frac{1}{20}\)CMR: E không phải là số tự nhiên
Tính bằng cách hợp lí nhất :
1/ \(\left(\frac{4}{9}-\frac{5}{11}\right):\frac{3}{10}+\left(\frac{3}{9}-\frac{9}{11}\right):\frac{3}{10}-\left(\frac{2}{9}-\frac{2}{8}\right).\frac{-10}{3}\)
2/ \(\frac{1}{2}.\frac{1}{-3}+\frac{1}{-3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{-5}+\frac{1}{-5}.\frac{1}{6}\)
3/ \(-\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
App giải toán không cần nhập đề chỉ cần chụp ảnh cho cả nhà đây: https://www.facebook.com/watch/?v=485078328966618
\((\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}).x=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+...+\frac{8}{2}+\frac{9}{1}\)
Ai bt làm thì giúp mk với!
Đặt \(A=\frac{1}{9}+\frac{2}{8}+...+\frac{8}{2}+\frac{9}{1}\)
\(\Rightarrow A=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+...+\frac{8}{2}+\left(1+1+...+1\right)\left(9cs1\right)\)
\(\Rightarrow A=\left(\frac{1}{9}+1\right)+\left(\frac{2}{8}+1\right)+...+\left(\frac{8}{2}+1\right)+1\)
\(\Rightarrow A=\frac{10}{9}+\frac{10}{8}+...+\frac{10}{2}+\frac{10}{10}\)
\(\Rightarrow A=10.\left(\frac{1}{2}+...+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
Mà \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{10}\right).x=A\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right).x=\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right).10\)
\(\Rightarrow x=10\)
Vậy \(x=10\)
1 CÁC BN TÍNH RÕ RÀNG NHA MK SẼ TÍCK CHO NHA !
\(\frac{1}{2}-\frac{-3}{6}+\frac{5}{3}-\frac{9}{12}\)
\(\frac{-2}{3}.\frac{3}{11}+\frac{-16}{9}:\frac{11}{3}\)
\(\frac{^{2^0}}{3}-\sqrt{9}+-\frac{1^2}{2}\)
Ta có:
\(=\frac{6}{12}+\frac{6}{12}+\frac{20}{12}-\frac{9}{12}\)
\(=\frac{29}{12}\)
Hok tốt
\(\frac{10+\frac{9}{2}+\frac{8}{3}+\frac{7}{4}+ \frac{6}{5}+\frac{5}{6}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}+\frac{1}{11}}\)