\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{1997.1999}-\frac{1}{1999.2001}\)

     \(=\frac{1}{1.3}-\frac{1}{1999.2001}\)

       Bạn tính kết quả nhé

a)\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

=\(\frac{9.4}{1.3.5}+\frac{9.4}{3.5.7}+\frac{9.4}{5.7.9}+...+\frac{9.4}{25.27.29}\)

=\(9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

=\(9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

=\(9.\left(\frac{1}{3}-\frac{1}{27.29}\right)=9.\left(\frac{1}{3}-\frac{1}{783}\right)=9.\left(\frac{261}{783}-\frac{1}{783}\right)=9.\frac{260}{783}\)

=\(\frac{260}{87}\)

b)

ta có: \(3=\frac{261}{87}>\frac{260}{87}\)

vậy A<3

KẾT BẠN VỚI MÌNH NHÉ NGƯỜI ĐẸP

mk cung dinh hoi bai nay

$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$

Xét tử số có dạng : \(\frac{1}{\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)}=\frac{1}{4}\left[\frac{1}{\left(2n+1\right)\left(2n+2\right)}-\frac{1}{\left(2n+2\right)\left(2n+3\right)}\right]\) với \(n\in N\)

Ta có : \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2005.2007.2009}\)

\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\frac{1}{4}.\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\frac{1}{4}\left(\frac{1}{5.7}-\frac{1}{7.9}\right)+...+\frac{1}{4}\left(\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)

\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{2007.2009}\right)\)

Xét mẫu số có dạng : \(\frac{1}{\left(2n+1\right)\sqrt{2n+3}+\left(2n+3\right)\sqrt{2n+1}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}=\frac{1}{2}.\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với  \(n\in N\)

Áp dụng : \(\frac{1}{1\sqrt{3}+3\sqrt{1}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{2007\sqrt{2009}+2009\sqrt{2007}}\)

\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2009}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)\)

Suy ra : \(M=\frac{\frac{1}{4}\left(\frac{1}{3}-\frac{1}{2007.2009}\right)}{\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)}\)

Tới đây bài toán đã gọn hơn , bạn tự tính nhé :)

1)\(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\)

\(\Rightarrow4S=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{2003.2005.2007}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2003.2005}-\frac{1}{2005.2007}\)

\(=\frac{1}{3}-\frac{1}{4024035}=\frac{1341345}{4024035}=\frac{1}{3}\)

\(\Rightarrow S=\frac{1}{3}:4\approx0,08\)

2)\(S=\frac{1}{3}:4=\frac{1}{12}\)

\(S=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2003.2005.2007}\)

\(S=\frac{2}{2}.\frac{1}{1.3.5}+\frac{2}{2}.\frac{1}{3.5.7}+\frac{2}{2}.\frac{1}{5.7.9}+...+\frac{2}{2}.\frac{1}{2003.2005.2007}\)

\(S=\frac{1}{2}.\frac{2}{1.3.5}+\frac{1}{2}.\frac{2}{3.5.7}+\frac{1}{2}.\frac{2}{5.7.9}+...+\frac{1}{2}.\frac{2}{2003.2005.2007}\)

\(S=\frac{1}{2}.\left(\frac{2}{1.3.5}+\frac{2}{3.5.7}+\frac{2}{5.7.9}+...+\frac{2}{2003.2005.2007}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2003.2005}-\frac{1}{2005.2007}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{2005.2007}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{4024035}\right)=\frac{1}{2}.\frac{1341345}{4024035}=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

Vậy \(S=\frac{1}{6}\)

\(2A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right).2\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)

\(A=\frac{98}{99}:2\)

\(A=\frac{49}{99}\)

Ta có:

\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

\(\Rightarrow A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)

\(\Rightarrow A=9.\frac{1}{3}-9.\frac{1}{783}\)

\(\Rightarrow A=3-\frac{1}{87}\)

Vì \(3-\frac{1}{87}< 3.\)

\(\Rightarrow A< 3\left(đpcm\right).\)

Chúc bạn học tốt!