a) \(\dfrac{1}{9.13}+\dfrac{1}{13.17}+...+\dfrac{1}{\left(4x+1\right)\left(4x+5\right)}=\dfrac{10}{441}\)
\(\dfrac{1}{4}.\left(\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+...+\dfrac{1}{4x+1}-\dfrac{1}{4x+5}\right)=\dfrac{10}{441}\)
\(\dfrac{1}{4}.\left(\dfrac{1}{9}-\dfrac{1}{4x+5}\right)=\dfrac{10}{441}\)
\(\dfrac{1}{9}-\dfrac{1}{4x+5}=\dfrac{10}{441}.4\)
\(\dfrac{1}{4x+5}=\dfrac{1}{9}-\dfrac{40}{441}\)
\(\dfrac{1}{4x+5}=\dfrac{1}{49}\)
\(4x+5=49\)
\(4x=49-5\)
\(4x=44\)
\(x=44:4\)
\(x=11\)
c: \(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{26}{27}\)
=>\(\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{26}{27}\)
=>\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{13}{27}\)
=>\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{13}{27}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{13}{27}\)
=>\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{13}{27}\)
=>\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{13}{27}=\dfrac{1}{54}\)
=>x+1=54
=>x=53(nhận)
d: \(\left(\dfrac{1}{1\cdot101}+\dfrac{1}{2\cdot102}+...+\dfrac{1}{10\cdot110}\right)x=\dfrac{1}{1\cdot11}+\dfrac{1}{2\cdot12}+...+\dfrac{1}{100\cdot110}\)
=>\(\dfrac{1}{100}\left(\dfrac{100}{1\cdot101}+\dfrac{100}{2\cdot102}+...+\dfrac{100}{10\cdot110}\right)\cdot x=\dfrac{1}{10}\left(\dfrac{10}{1\cdot11}+\dfrac{10}{2\cdot12}+...+\dfrac{10}{100\cdot110}\right)\)
=>\(\dfrac{1}{100}\left(1-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+...+\dfrac{1}{10}-\dfrac{1}{110}\right)\cdot x=\dfrac{1}{10}\left(\dfrac{1}{1}-\dfrac{1}{11}+\dfrac{1}{2}-\dfrac{1}{12}+...+\dfrac{1}{100}-\dfrac{1}{110}\right)\)
=>\(x\cdot\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{101}-\dfrac{1}{102}-...-\dfrac{1}{110}\right)=10\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}-\dfrac{1}{11}-\dfrac{1}{12}-...-\dfrac{1}{110}\right)\)
=>\(x\cdot\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{101}-\dfrac{1}{102}-...-\dfrac{1}{110}\right)=10\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}-\dfrac{1}{101}-\dfrac{1}{102}-...-\dfrac{1}{110}\right)\)
=>x=10
