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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow x+2=41\)

\(\Rightarrow x=41-2\)

\(\Rightarrow x=39\)

Vậy x = 39

A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)

\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)

\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)

X=16

17 - 1= 16

= > x = 16

 tk mình nha

17 - 1= 16

= > x = 16

 tk mình nha

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}\right)=2.\frac{20}{41}\) 

\(\Rightarrow\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\) 

\(\Rightarrow\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{x}-\frac{1}{x+2}\right)=\frac{40}{41}\) 

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

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

\(\Rightarrow x+2=41\) 

\(\Rightarrow x=39\) 

Vậy x = 39

Nhớ t.i.c.k cho mình nha!

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

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

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

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

\(A=\frac{x}{2x+1}\) 

Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)

x=49

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right)\left(2x-1\right)}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}+\frac{1}{2x+1}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)

\(\Rightarrow99x=49\left(2x+1\right)\)

\(\Rightarrow99x=98x+49\)

\(\Rightarrow x=49\)

Vậy : \(x=49\)

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

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

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

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

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

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

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

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

=> không có kết quả

\(a)\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}\right)=2\cdot\frac{20}{41}\)

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x+1=40\\x+2=41\end{cases}\Leftrightarrow\hept{\begin{cases}x=40-1\\x=41-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=39\\x=39\end{cases}}}\)

Vậy x=39

\(b)|x+2016|\ge0\forall x;|x+2017|\ge0\forall x\)

\(\Leftrightarrow x+2016+x+2017+2018=3x\)

\(\Leftrightarrow2x+6051=3x\)

\(\Leftrightarrow6051=3x-2x\)

\(\Leftrightarrow6051=x\)

Vậy x=6051

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

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

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

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

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

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

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

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

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

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