Question

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

Answer 1

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

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

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

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

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

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

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

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

Answer 2

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

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

Answer 3

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!