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\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{11}{11}-\frac{1}{11}\)
\(=\frac{10}{11}\)
Chúc bạn học tốt !!!
Bài làm:
Ta có: \(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)
\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)\(\Rightarrow\frac{2}{5}< S\)
Cái còn lại tự CM
A= 1/2.2 + 1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + 1/7.7 + 1/8.8 + 1/9.9
Vì 1/2.2 > 1/2.3; 1/3.3 > 1/3.4 ; 1/5.5 > 1/5.6;...... nên
1/2.2 +1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + 1/7.7 + 1/8.8 + 1/9.9 > 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10
Ta có: 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10
= 1/2-1/3 + 1/3 -1/4 + 1/4-1/5+...+1/9-1/10
= 1/2- 1/10
= 2/5
Vì A < 2/5 mà 2/5 <7/8 nên 2/5 < A < 7/8
Vậy....
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\Rightarrow A< \frac{8}{9}\)(1)
Lại có \(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\Rightarrow A>\frac{2}{5}\)(2)
Từ (1) (2) => \(\frac{2}{5}< A< \frac{8}{9}\left(\text{ĐPCM}\right)\)
Bài làm :
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>\frac{1}{2}-\frac{1}{10}\)
\(A>\frac{2}{5}\left(1\right)\)
Ta cũng có :
\( A=\frac{1}{2.2}+\frac{1}{3.3}+......+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-......+\frac{1}{8}-\frac{1}{9}\)
\(A< 1-\frac{1}{9}\)
\(A< \frac{8}{9}\left(2\right)\)
\(\text{Từ (1) và (2) }\Rightarrow\frac{2}{5}< A< \frac{8}{9}\)
=> Điều phải chứng minh
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
=\(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
=\(2\left(1-\frac{1}{21}\right)\)
=\(\frac{2.20}{21}=\frac{40}{21}\)
2/1.3+2/3.5+2/5.7+...+2/101.103=1/1-1/3+1/3-1/5+1/5-1/7+...+1/101-1/103
=1-1/103=102/103
k cho mình nhé chắc chắn đúng
\(\frac{33}{2}+\frac{33}{6}+\frac{33}{18}+\frac{33}{54}+\frac{33}{162}+\frac{33}{486}\)
\(=\frac{33.3+33.3+33.3+33.3+33.3}{486}\)
\(=\frac{99.5}{486}\)
\(=\frac{495}{486}\)
Gọi \(A=\frac{33}{2}+\frac{33}{6}+...+\frac{33}{486}\)
\(A=33.\left[\left(\frac{1}{1.2}+\frac{1}{2.3}\right)+\left(\frac{1}{3.6}+\frac{1}{6.9}\right)\left(\frac{1}{9.18}+\frac{1}{18.27}\right)\right]\)
\(A=33.\left[\frac{2}{3}+\frac{2}{9}+\frac{2}{27}\right]\)
\(A=66.\left[\frac{9}{27}+\frac{3}{27}+\frac{1}{27}\right]\)
\(A=66.\frac{13}{27}\)
\(A=\frac{286}{9}\)
sai hay đúng cx ko biết nha
bai nay de thui
nhung bay gio mk ban
luc nao ranh mk lam
cho nha
minhpham@gmail.com
A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/2001-1/2003+1/2003-1/2005
A=1-1/2005
A=2004/2005
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