Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A= \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ ... + \(\frac{1}{19.20.21}\)< \(\frac{1}{4}\)
= 1 - \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\)
= 1 - ( \(\frac{1}{2}-\frac{1}{3}\)+ \(\frac{1}{2}-\frac{1}{3}\)) + ... + ( \(\frac{1}{19}-\frac{1}{20}+\frac{1}{19}-\frac{1}{20}\)) - \(\frac{1}{21}\)
= 1 - \(\frac{1}{21}\)
= \(\frac{20}{21}\)< \(\frac{1}{4}\)
=> Đề bài có sai ko bạn?
\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right).x=5\)
\(\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{21-19}{19.20.21}\right).x=5\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right).x=5\)
\(\left(\frac{1}{1.2}-\frac{1}{20.21}\right).x=5\)
\(\frac{209}{420}.x=5\)
\(\Rightarrow x=5\div\frac{209}{420}=\frac{2100}{209}\)
\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right).x=5\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}\right).2.x=5\)
\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{19.20}-\frac{1}{20.21}\right)\right).x.2=5\)
\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\right).x=5\div2\)
\(\left(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{20.21}\right)\right).x=2,5\)
\(\left(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{420}\right)\right).x=2,5\)
\(\left(\frac{1}{2}\times\frac{209}{420}\right)\times x=2,5\)
\(\frac{209}{840}\times x=2,5\)
\(x=2,5\div\frac{209}{840}=10\frac{10}{209}\)
Có \(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)
\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)
...
\(\frac{1}{17.18.19}=\frac{1}{2}\left(\frac{1}{17.18}-\frac{1}{18.19}\right)\)
=>\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{17.18.19}\)=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{17.18}-\frac{1}{18.19}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{18.19}\right)=\frac{1}{2}.\frac{1}{2}-\frac{1}{2}.\frac{1}{18.19}< \frac{1}{4}\)
= 1/2.(2/1.2.3+2/2.3.4+.....+2/50.51.52
=1/2.(1/1.2-1/2.3+1/2.3-1/3.4+....+1/50.51-1/51.52
=1/2.(1/1.2-1/51.52)
=1/2.(1/2-1/2652)
=1/2.1325/2652
=1325/5304
A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/50.51-1/51.52
A=1/1.2-1/51.52
phần còn lại tự giải nhé
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+..........+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=\frac{1}{1.2}-\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{90}\)
\(=\frac{45}{90}-\frac{1}{90}\)
\(=\frac{44}{90}\)
\(=\frac{22}{45}\)
Đặt A=(đã cho).
=>2A=2/1*2*3+2/2*3*4+2/3*4*5+...+2/37*38*39.
=>2A=1/1*2-1/2*3+1/2*3-1/3*4+...+1/37*38-1/38*39.
=>2A=1/1*2=1/38*39.
Đến đây tự bấm máy nha.
tk mk nha.
chắc chắn đúng,nay mk làm bài này.
-chúc ai tk mk học giỏi-
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2004.2005.2006}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2004.2005}-\frac{1}{2005.2006}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2005.2006}\right)\)
\(=\frac{1}{4}-\frac{1}{2.2005.2006}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{420}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)
=\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{21-19}{19.20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{420}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)