\(1\times2+1\times2\times3+...+1\times2\times3\times...\times100\)
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
\(1\times2\times3\times.........\times9-1\times2\times3\times...........\times8-1\times2\times3\times........8^2\)
Giúp mình nhé ai nhanh nhất mình tick
\(1\cdot2\cdot3\cdot...\cdot8\cdot9-1\cdot2\cdot3\cdot...\cdot8-1\cdot2\cdot3\cdot...\cdot8^2\)
=\(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot\left(9-1-8\right)\)(đặt 1*2*3*...*8 ra ngoài)
=\(1\cdot2\cdot3\cdot...\cdot8\cdot0=0\)
\(A=\left[1-\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+......+\frac{1}{98\times99\times100}\right)\right]\times\frac{14851}{19800}\)
\(y=\frac{1\times100+2\times99+3\times98...+99\times2+100\times1}{1\times2+2\times3+3\times4+...+99\times100+100\times101}=?\)
Cho B= \(\frac{1\times2}{1\times2\times3}+\frac{1\times2}{1\times2\times4}+\frac{1\times2}{1\times2\times3\times4}+\frac{1\times2}{1\times2\times3\times4\times5}+....+\frac{1\times2}{n,giao}\left(n\in N,n\ge3\right)\)
chứng tỏ B nhỏ hơn 3
Tính \(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{98\times99\times100}\)
sud kênh Mik ủng hộ với tên kênh là M.ichibi
kênh làm về MINECRAFT
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
tự tính
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{99\cdot99}-\frac{1}{99\cdot100}\)
\(2A=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}=\frac{4949}{9900}\Rightarrow A=\frac{4949}{19800}\)
tính
a, \(1\times2\times3\times...\times2018-1\times2\times3\times...\times2017^2\)
b,\(1500-\left\{5^2\times2^3-11\times\left[7^2-5\times2^3+8\times\left(11^2-121\right)\right]\right\}\)
Tính nhanh:\(\frac{1+2}{1\times2}+\frac{1+2+3}{1\times2\times3}+...+\frac{1+2+...+999}{1\times2\times...\times999}+\frac{1+2+...+999+1000}{1\times2\times...\times999\times1000}\)
Đề bài có vẻ bất ổn em ơi?
Tính nhanh:\(\frac{1\times2}{1+2}+\frac{1\times2\times3}{1+2+3}+..+\frac{1\times2\times...\times999}{1+2+...+999}+\frac{1\times2\times...\times999\times1000}{1+2+...+999+1000}\)
1x2/1+2 + ... + 1x2x ... x 999x1000/1+2+ ... +1000
= 1 + ... + 1
= 1 x 1000
= 1000