Tính nhanh \(A=\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{31.33.35}\)
b) Giúp mik ra một bài tương tự
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PT
7 tháng 11 2017
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{1997.1999}-\frac{1}{1999.2001}\)
\(=\frac{1}{1.3}-\frac{1}{1999.2001}\)
Bạn tính kết quả nhé
L
0
MT
11 tháng 7 2015
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
TS
3 tháng 10 2015
\(2A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right).2\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)
\(A=\frac{98}{99}:2\)
\(A=\frac{49}{99}\)
ST
2
XO
27 tháng 8 2019
\(\frac{20}{1.3.5}+\frac{20}{3.5.7}+\frac{20}{5.7.9}+...+\frac{20}{25.27.29}\)
\(=5.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=5.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=5.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(=5.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(=5.\frac{260}{783}\)
\(=\frac{1300}{783}\)
27 tháng 8 2019
Ta có:\(\frac{1}{\left(n-2\right)n}-\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)-\left(n-2\right)n}{\left(n-2\right)n\cdot n\left(n+2\right)}\)
\(=\frac{n\left(n+2-n+2\right)}{n\cdot\left(n-2\right)n\left(n+2\right)}=\frac{4}{\left(n-2\right)n\left(n+2\right)}\)
Áp dụng\(\frac{20}{1.3.5}+\frac{20}{3.5.7}+...+\frac{20}{25.27.29}\)
\(=5\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=5\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=5\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(=5\cdot\frac{261-1}{783}=5\cdot\frac{260}{783}=\frac{1300}{783}\)
24 tháng 4 2019
\(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+\frac{36}{5\cdot7\cdot9}+...+\frac{36}{25\cdot27\cdot29}\)
\(=9\left[\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{25\cdot27\cdot29}\right]\)
\(=9\left[\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right]\)
\(=9\left[\frac{1}{3}-\frac{1}{783}\right]=9\cdot\frac{260}{783}=\frac{260}{87}\)
KN
24 tháng 4 2019
Đặt \(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{261}{783}-\frac{1}{783}\)
\(\Rightarrow\frac{1}{9}A=\frac{260}{783}\)
\(\Rightarrow A=\frac{260}{783}\div\frac{1}{9}\)
\(\Rightarrow A=\frac{2340}{783}=\frac{260}{87}\)
20 tháng 3 2016
$\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 $\Rightarrow B<3$
a) \(A=\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{31.33.35}\)
=>\(2A=\frac{2}{3.5.7}+\frac{2}{5.7.9}+...+\frac{2}{31.33.35}\)
\(=\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{31.33}-\frac{1}{33.35}\)
\(=\frac{1}{15}-\frac{1}{1155}=\frac{77}{1155}-\frac{1}{1155}=\frac{76}{1155}\)
=> \(A=\frac{76}{1155}:2=\frac{76}{1155}.\frac{1}{2}=\frac{38}{1155}\)
b)
\(B=\frac{1}{1.4.7}+\frac{1}{7.10.13}+...+\frac{1}{54.57.60}\)