Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\times\frac{98}{99}\)
n=\(\frac{196}{297}\)
Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé.
\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)
\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)
\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)
Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{100-98}{98.99.100}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{18000}\)
=1-/2-1/3+1/2-1/3-1/4+1/5-1/6-1/7+1/6-1/7-1/8-.........-1/98-1/99-1/100
=1-1/100
=99/100
2/1*2*3+2/3*4*5+...+2/2009*2010*2011
A=2/2*(1/1-1/2-1/3+1/2-1/3-1/4+1/4-1/5-1/6+...+1/2009-1/2010-1/2011
A=1*(1-1/2011)
A=1*2010/2011=2010/2011
suy ra: 2010/2011<1
suy ra 1/2 của 1 lớn hơn 2010/2011
VẬY A NHỎ HƠN 1/2
VẬY
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
Tính nhanh :
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{2}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\left(\frac{20}{20}-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\frac{98}{99}\)
\(A=\frac{49}{99}\)
=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
=1-\(\frac{1}{99}\)
=\(\frac{98}{99}\)
Sai đề rồi bạn. Phải là thế này chứ:
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{100}=\frac{1}{6}+\frac{1}{2}-\frac{1}{100}=\frac{197}{300}\)
Lời giải:
$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2014-2012}{2012.2013.2014}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{2012.2013}-\frac{1}{2013.2014}$
$=\frac{1}{1.2}-\frac{1}{2013.2014}< \frac{1}{2}$
$\Rightarrow A< \frac{1}{2}:2$
Hay $A< \frac{1}{4}$
câu M hình như đề sai rồi bn ơi
câu m bị sai đề ở chỗ 2 phần 99 nhân 100 ý