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7 tháng 9 2017

Đặt \(A=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^3}{5.7}+...+\dfrac{1006^2}{2011.2013}\)

\(\Rightarrow4A=\dfrac{4.1^2}{1.3}+\dfrac{4.2^2}{3.5}+\dfrac{4.3^3}{5.7}+...+\dfrac{4.1006^2}{2011.2013}\)

\(\Rightarrow4A=1006+\dfrac{1}{2}.\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{2011}-\dfrac{1}{2013}\right]\)

\(\Rightarrow A=\dfrac{1006+\dfrac{1}{2}.\left(1-\dfrac{1}{2013}\right)}{4}\)

\(\Rightarrow A=251,6249\)

8 tháng 9 2017

Cho mik hỏi tại sao lại tính được đến 1006+1/2(1-1/3....) vấn đề của mik là số 1006 ý giải rõ giúp mik nha . Cảm ơn nhiều!

6 tháng 4 2017

\(\sum\limits^{2016}_{x=1}\left(\dfrac{x^2}{\left(2x-1\right)\left(2x+1\right)}\right)\)

8 tháng 4 2017

bạn có thể giải thích rõ dùm mk ko

19 tháng 1 2018

Dat A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\)

2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\)

= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-....+\dfrac{1}{13}-\dfrac{1}{15}\)

= 1-\(\dfrac{1}{15}=\dfrac{14}{15}\)

=> A=\(\dfrac{7}{15}\)

Ta co : \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

=> \(\dfrac{7}{15}x-\dfrac{7}{15}+\dfrac{7}{15}=\dfrac{3}{5}x\)

=> \(\dfrac{7}{15}x-\dfrac{3}{5}x=0\)

=> x\(\left(\dfrac{7}{15}-\dfrac{3}{5}\right)=0\)

=> x\(\left(-\dfrac{2}{15}\right)=0\)

=> x=0

19 tháng 1 2018

\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=> \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{7}{15}x-\dfrac{7}{15}=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{7}{15}x-\dfrac{3}{5}x=\dfrac{-7}{15}+\dfrac{7}{15}\)

<=> \(\dfrac{-2}{15}x=0\)

<=> \(x=0\)

Vậy: \(s=\left\{0\right\}.\)

19 tháng 1 2018

xong r. k cần nữa nhé

17 tháng 1 2018

2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\)

=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2007}-\dfrac{1}{2009}\)

= 1- \(\dfrac{1}{2009}\)

= \(\dfrac{2008}{2009}\)

=> S=\(\dfrac{1004}{2009}\)

10 tháng 5 2018

làm sao ra được \(\dfrac{1004}{2009}\)

16 tháng 5 2018

1)\(2a^4+1\ge2a^3+a^2\)

\(\Leftrightarrow2a^4-2a^3-a^2+1\ge0\)

\(\Leftrightarrow\left(a^4-2a^3+a^2\right)+\left(a^4-2a^2+1\right)\ge0\)

\(\Leftrightarrow\left(a^2-a\right)^2+\left(a^2-1\right)^2\ge0\)(luôn đúng)

"="<=>a=1

16 tháng 5 2018

Ta có:\(2A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{9\cdot11}\)

\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\)

\(2A=1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(B=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{9\cdot11}\right)\)

\(B=\dfrac{4}{1\cdot3}\cdot\dfrac{9}{2\cdot4}\cdot...\cdot\dfrac{100}{9\cdot11}\)

\(B=\dfrac{2\cdot2\cdot3\cdot3\cdot...\cdot10\cdot10}{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}\)

\(B=\dfrac{20}{11}\)

\(\Rightarrow11< 2x< 20\)

\(\Rightarrow x\in\left\{6;7;8;9\right\}\)

NV
25 tháng 3 2023

\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)

\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)

27 tháng 2 2018

\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(\Rightarrow2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)

\(\Rightarrow2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\) \(\Rightarrow2S=1-\dfrac{1}{2n+1}\)

\(\Rightarrow S=\dfrac{n}{2n+1}\)

27 tháng 2 2018

Ta có : \(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

ta được \(\dfrac{1}{1.3}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right);\dfrac{1}{3.5}=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right);\dfrac{1}{5.7}=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)

\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\) vậy \(S=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{n}{2n+1}\)

a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)