Tìm số nguyên tố \(n\) lớn nhất để: \(\left(1\times2\times3\times...\times97\times98\right)+\left(1\times2\times3\times...\times98\times99\times100\right)⋮n\)
Tính:
\(M=\frac{1\times2\times3\times4\times5\times6\times7\times8\times9\times...\times97\times98\times99}{10}\)
\(M=\frac{1.2.3.4.5...98.99}{10}\)
\(M=1.2.3.4.5.6.7.8.9.11.12...98.99\)
\(A=\left[1-\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+......+\frac{1}{98\times99\times100}\right)\right]\times\frac{14851}{19800}\)
\(y=\frac{1\times100+2\times99+3\times98...+99\times2+100\times1}{1\times2+2\times3+3\times4+...+99\times100+100\times101}=?\)
Cho \(S=1\times2\times3+2\times3\times4+3\times4\times5+...+97\times98\times99\)
Hãy tìm số tự nhiên n nhỏ nhất có thể để \(4\times S+n\) là một số chính phương.
tìm x biết
\(\frac{x\times\left(1\times2+2\times3+3\times4+...+98\times99\right)}{98\times100\times33}=2010-|-2011|\)
a,A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
b,B=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{998\times999\times100}\)
c,C=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1\times98+2\times97+3\times96+...+98\times1}\)
Tính
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99+100\right)}{\left(1\times100+2\times99+3\times98+...+99\times2+100\times1\right)\times2013}\)
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+.....+\left(1+2+3+4+......+100\right)}{\left(1.100+2.99+3.98+.......+99.2+100.1\right).2013}\)
\(=\frac{1.100+2.99+3.98+......+99.2+100.1}{\left(1.100+2.99+3.98+.....+99.2+100.1\right).2013}\)
\(=\frac{1}{2013}\)
a) Tìm M biết:
\(M=1\times2\times3\times4\times5\times...\times97\times98\times99\)
b) Tính: \(\frac{M}{10}\)= ?
Nhớ giải ra nhé!!!
tính
\(\frac{1}{1\times2\times3}-\frac{1}{2\times3\times4}-\)...\(-\frac{1}{97\times98\times99}\)
Ta có \(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)...... , \(\frac{1}{97.98}-\frac{1}{98.99}=\frac{2}{97.98.99}\)
vậy 2 xA = \(\frac{2}{1.2.3.}\) -\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)-.\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\).....-\(\frac{1}{97.98}\)+\(\frac{1}{98.99}\)
=1/3-1/6+1/(98.99) =1/6 +1/(98.99)
=> A = 1/12+\(\frac{1}{2.98.99}\)