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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow x+2=41\)

\(\Rightarrow x=41-2\)

\(\Rightarrow x=39\)

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

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

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

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

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

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

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

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

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

=> không có kết quả

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)

\(\frac{x}{14}=\frac{16}{21}\)

\(\Rightarrow x\cdot=21=14\cdot16\)

\(\Rightarrow x\cdot21=224\)

\(\Rightarrow x=\frac{224}{21}\)

bạn nào giải giúp mình với

nếu đúng thì mình sẽ ***

=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

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

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

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{40}{41}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

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

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

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

=> x+2=41

          x=41-2

          x=39

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{7}\)

\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{7}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{7}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{7}\)

\(\Rightarrow\frac{1}{x+2}=-\frac{9}{7}\)

\(\Rightarrow-9\left(x+2\right)=7\)

\(\Rightarrow x+2=-\frac{7}{9}\)

\(\Rightarrow x=-\frac{25}{9}\)

Vậy \(x=-\frac{25}{9}\)

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

x=\(\frac{4}{13}\)

b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)

Giống câu a

=> \(2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{16}{34}\)

=>\(2.\left(1-\frac{1}{x+2}\right)=\frac{16}{34}\)

=>\(1-\frac{1}{x+2}=\frac{4}{17}\)

=> \(\frac{1}{x+2}=\frac{13}{17}\)

=>\(x=-\frac{9}{13}\)