thì làm sao???Hỏi xong rồi tự trả lời thì có ích gì

(✿◠‿◠)(๛ČℌUƔÊŇ♥Ť❍Ą́Ňツ)

Ê nhóc đừng có nghĩ lung tung 

\(A=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)\cdot\cdot\cdot\left(1-\frac{2}{2011}\right)\)

\(A=\left(\frac{5-2}{5}\right)\left(\frac{7-2}{7}\right)\left(\frac{9-2}{9}\right)\cdot\cdot\cdot\left(\frac{2011-2}{2011}\right)\)

\(A=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot\cdot\cdot\frac{2009}{2011}\)(các thừa số trên tử giống dưới mẫu mình lượt bỏ đi nhé!)

\(A=\frac{3}{2011}\)

\(A=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)...\left(1-\frac{2}{2011}\right)\)

\(=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}...\frac{2009}{2011}\)

\(=\frac{3}{2011}\)

\(A=\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{2}{9}\right).....\left(1-\frac{2}{2011}\right)\)

\(A=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}........\frac{2009}{2011}\)

\(A=\frac{3}{2011}\)

Ta có : \(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right).....\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)

\(=\dfrac{5}{3}.\dfrac{7}{5}....\dfrac{2013}{2011}.\dfrac{2015}{2013}=\dfrac{2015}{3}\)
 

\(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right)\left(1\dfrac{2}{7}\right)...\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)

\(=\dfrac{5}{3}.\dfrac{7}{5}.\dfrac{9}{7}.....\dfrac{2013}{2011}.\dfrac{2015}{2013}\)

\(=\dfrac{2015}{3}\)

\(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right)\left(1\dfrac{2}{7}\right)...\left(1\dfrac{2}{2013}\right)\)

\(=\dfrac{5}{3}.\dfrac{7}{5}.\dfrac{9}{7}...\dfrac{2015}{2013}=\dfrac{2015}{3}\)

\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)

\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)

(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)

=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10

=8/15.48/63.120/143.3/8.35/48.9/10

=384/945.360/1144.315/480

=138240/1081080.315/480

=43545600/518918400=84/1001

khó quá

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\)    \(\left(1-\frac{1}{11}\right)\)\(\left(1-\frac{1}{13}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\)
 \(=1.\left(\frac{1}{3}-\frac{1}{2}\right).\left(\frac{1}{5}-\frac{1}{4}\right).\left(\frac{1}{7}-\frac{1}{6}\right)\)\(.\left(\frac{1}{11}-\frac{1}{10}\right).\frac{1}{13}\)
  \(=1.1.1.1.1.\frac{1}{13}\)
   \(=\frac{1}{13}\)
Cái này là chỉ do mình nghĩ thôi, ko biết đúng ko nữa
Có sai cko mình sorry nha!
Tk và kb hộ mình  với! thanks 

C=(1+2/3).(1+2/5).(1+2/7)......(1+2/2009).(1+2/2011)

C=5/3.7/5.9/7......2011/2009.2013/2011

C=5.7.9.....2013/3.5.7.....2009.2011

C=2013/3

\(=\left(-1\right).1.\left(-1\right).1.......\left(-1\right).1=1\)

\(\left(-1\right).\left(-1\right)^2.\left(-1\right)^3.\left(-1\right)^4...........\left(-1\right)^{99}.\left(-1\right)^{100}\)

\(=\left(-1\right).1.\left(-1\right).1.......\left(-1\right).1\)

\(=1\)

C=(67/111+2/33-15/117).(1/3-1/4-1/12)

C=(67/111+2/33-15/117).0

C=0

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