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9 tháng 5 2019

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{x(x+2)}=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x(x+2)}\right]=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right]=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}\left[1-\frac{1}{x+2}\right]=\frac{20}{41}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\Leftrightarrow x+2=41\Leftrightarrow x=39\)

9 tháng 5 2019

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}.\)

\(1-\frac{1}{x+2}=\frac{20}{41}\Rightarrow\frac{1}{x+2}=\frac{21}{41}=\frac{21}{21x+42}\Rightarrow21x+42=41\Rightarrow x=-\frac{1}{21}\)

13 tháng 8 2019

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)

\(\Leftrightarrow20\left(x+2\right)=41\)

\(\Leftrightarrow x-2=\frac{41}{20}\)

\(\Leftrightarrow x=\frac{41}{20}+2\)

\(\Leftrightarrow x=\frac{81}{20}\)

13 tháng 8 2019

\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\) 

\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)

8 tháng 4 2016

bạn nào giải giúp mình với

nếu đúng thì mình sẽ ***

8 tháng 4 2016

=1/2*(1-1/3+1/3-1/5+....+1/x+1/x+2)

=1/2*(1-1/x+2)

=>1/2*x+1/x+2=20/21

Đến đó đưa về giống tìm x nha

25 tháng 3 2016

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{5}{11}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{5}{11}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{5}{11}\Rightarrow1-\frac{1}{x+2}=\frac{5}{11}\div\frac{1}{2}=\frac{10}{11}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{10}{11}=\frac{1}{11}\Rightarrow x+2=11\Rightarrow x=11-2=9\)

\(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{x+\left(x+2\right)}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{x}-\frac{1}{x+2}\)

\(=1-\frac{1}{x+2}=\frac{5}{11}\)

\(\frac{1}{x+2}=1-\frac{5}{11}=\frac{6}{11}\)

=> không có kết quả

17 tháng 4 2016

\(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+........+\frac{x}{39\cdot41}=\frac{1}{41}\)

\(\Rightarrow x\cdot\left[\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.........+\frac{1}{39\cdot41}\right)\right]=\frac{1}{41}\)

\(\Rightarrow x\cdot\left[\frac{1}{2}\cdot\left(1-\frac{1}{41}\right)\right]=\frac{1}{41}\)

\(\Rightarrow x\cdot\left(\frac{1}{2}\cdot\frac{40}{41}\right)=\frac{1}{41}\)

\(\Rightarrow x\cdot\frac{20}{41}=\frac{1}{41}\)

\(\Rightarrow x=\frac{1}{41}:\frac{20}{21}\)

\(\Rightarrow x=\frac{1}{41}\cdot\frac{21}{20}\)

\(\Rightarrow x=\frac{21}{820}\)

ai k mh mh k lại

k cho mh nha

8 tháng 4 2015

Đặt   \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

       \(2A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\right)\)

       \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2x}{x\left(x+2\right)}=\frac{20}{41}\)

    Ta Có  \(\frac{1}{n}-\frac{1}{n+k}=\frac{n+k}{n\left(n+k\right)}-\frac{n}{n\left(n+k\right)}=\frac{k}{n\left(n+k\right)}\)

Áp Dụng Công Thức Trên Ta Có

      \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2x}{x\left(x+2\right)}=\frac{20}{41}\)

     \(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)

       \(2A=\frac{1}{1}-\frac{1}{x+1}=\frac{20}{41}\)

       \(\Rightarrow\frac{1}{x+1}=\frac{1}{1}-\frac{20}{41}\) 

       \(\Rightarrow\frac{1}{x+1}=\frac{21}{41}\)

       \(\Rightarrow\left(x+1\right).21=41\)

        \(\Rightarrow\left(x+1\right)=\frac{41}{21}\)

        \(\Rightarrow x=\frac{20}{21}\)

8 tháng 4 2015

\(1-\frac{1}{x+2}=\frac{20}{41}\Rightarrow\frac{1}{x+2}=\frac{21}{41}\Rightarrow x=mấybạntựtính\)

hơi khó đó tick mình nha Hoàng Thu Hà