Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=\frac{101}{120}+\frac{1}{2.3}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{19-18}{18.19}+\frac{20-19}{19.20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{20}\right)=\frac{101}{120}+\frac{19}{120}=\frac{120}{120}=1\)
a. Vì
1/2<2/3
3/4<4/5
.........
99/100<100/101 nên M<N
b.M.N=\(\frac{1.2.3.4......100}{2.3.4.5......101}\)=\(\frac{1}{101}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x+1 = 100
=> x = 100 - 1
=> x = 99
giải
=> 1/2-1/3+1/3-1/4+...+1/x-1/x-1 = 2/5
=> 1/2 -1/x-1 = 2/5
=> 1/x-1=1/2 - 2/5
=> 1/x-1=1/10
=> 1.10=1.(x-1)
=> x-1 = 10
=> x = 11
chúc bn may mắn
= 2.(1 / 2.3 + 1 / 3.4 + ..... + 1 / x (x + 1) = 2007/2009
= 2.(1/2 - 1/3 + 1/3 - +.......+ 1/x - 1/x+1) = 2007/2009
= 2.( 1/2 - 1/x+1) = 2007/2009
= 1 - 1/x+1 =2007/2009
= 1/x+1 = 1/2009
=> x + 1 = 2009
=> x = 2008
Ta có: 2/2.3 + 2/3.4 + .... + 2/x.(x+1) = 2007/2009
=> 2.[1/2.3+1/3.4+.....+1/x.(x+1)]=2007/2009
=> 2.(1/2-1/3+1/3-1/4 + .... + 1/x - 1/x+1) = 2007/2009
=> 2.(1/2-1/x+1)=2007/2009
=>1/2 - 1/x+1 = 2007/2009 : 2
=> 1/2 - 1/x+1 = 2007/4018
=> 1/x+1 = 2007/4018 +1/2
=> 1/x+1 =