Các câu hỏi tương tự
\(\frac{m}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......-\frac{1}{1334}+\frac{1}{1335}\) CM m là bội của 2003
chung to A=\(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2013!}<2\)
(n!=1.2.3....n duoc goi la giai thua)
Cho M =\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}vaN=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
a) Tinh tich M.N
b) chung minh M<N
c) Chung minh M < \(\frac{1}{10}\)
Bài 1 :
1 . Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
2 . Biết : 13 + 23 + ... + 103 = 3025
Tính : S = 23 + 43 + 63 + .... + 203
B=\(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
B=\(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(-1\frac{1}{5}.\frac{4.\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
Tính
\(B=-1\frac{1}{5}\)\(.\frac{4.\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
Cho n thuoc N*, chung to rang \(\frac{1}{1^2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+....+ \(\frac{1}{n^2}\)khong phai la mot so tu nhien
Help me!!!!!!!!!!