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2/1.3+2/3.5+...+2/x(x+2)= 40/41
1-1/3+1/3-1/5+...+1/x-1/(x+2)=40/41
1-1/(x+2)=40/41
1/(x+2)=1-40/41=1/41
x+2=41
x=41-2=39
Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)
\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)
\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)
Suy ra: x+2=82
hay x=80
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow20\left(x+2\right)=41\)
\(\Leftrightarrow x-2=\frac{41}{20}\)
\(\Leftrightarrow x=\frac{41}{20}+2\)
\(\Leftrightarrow x=\frac{81}{20}\)
\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)
Ta có :
\(\dfrac{1}{2}\)(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)
\(\dfrac{1}{2}\)(\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)
\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\)=\(\dfrac{40}{41}\)
\(\dfrac{1}{x+2}\)=\(\dfrac{1}{3}\)-\(\dfrac{40}{41}\)
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1