tính nhanh
\(\frac{3}{2.6}+\frac{3}{6.10}+\frac{3}{10.14}\)
\(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}\)
\(\frac{5}{2.4.6}+\frac{5}{4.6.8}+\frac{5}{6.8.10}\)
K
Khách
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DD
A = \(\frac{2.4.6+4.6.8+6.8.10+8.10.12+...+198.200.202}{1.3.5+3.5.7+5.7.9+7.9.11+...+97.99.101}\) =?
0
NV
2
8 tháng 8 2016
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 :)
NV
1
NL
2
8 tháng 5 2018
\(A=\frac{2}{2.4.6}+\frac{2}{4.6.8}+\frac{2}{6.8.10}+\frac{2}{8.10.12}\)
\(A=\frac{2}{48}+\frac{2}{192}+\frac{2}{480}+\frac{2}{960}\)
\(A=\frac{1}{24}+\frac{1}{96}+\frac{1}{240}+\frac{1}{480}\)
\(A=\frac{20}{480}+\frac{5}{480}+\frac{2}{480}+\frac{1}{480}\)
\(A=\frac{7}{120}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3
A = \(\dfrac{2}{2.4.6}\) + \(\dfrac{2}{4.6.8}\) + \(\dfrac{2}{6.8.10}\) + \(\dfrac{2}{8.10.12}\)
A = \(\dfrac{2}{2}\).(\(\dfrac{2}{2.4.6}\) + \(\dfrac{2}{4.6.8}\) + \(\dfrac{2}{6.8.10}\) + \(\dfrac{2}{8.10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{2.2}{2.4.6}\) + \(\dfrac{2.2}{4.6.8}\) + \(\dfrac{2.2}{6.8.10}+\dfrac{2.2}{8.10.12}\))
A = \(\dfrac{1}{2}\).( \(\dfrac{4}{2.4.6}+\dfrac{4}{4.6.8}+\dfrac{4}{6.8.10}+\dfrac{4}{8.10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{4.6}\) +\(\dfrac{1}{4.6}\) - \(\dfrac{1}{6.8}\) + \(\dfrac{1}{6.8}\) - \(\dfrac{1}{8.10}\) + \(\dfrac{1}{8.10}\) - \(\dfrac{1}{10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{8}-\dfrac{1}{120}\))
A = \(\dfrac{1}{2}\).\(\dfrac{7}{60}\)
A = \(\dfrac{7}{120}\)
R
chứng tỏ rằng : A=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+....+\frac{36}{25.27.29}< 3\)
1
VM
6 tháng 2 2020
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!
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 $\Rightarrow B<3$
DD
0
\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)
= \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)
= \(\dfrac{9}{28}\)
B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)
B = \(\dfrac{20}{63}\)