Tìm x, biết 1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/1540

a, \(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+...+\(\dfrac{1}{x\left(x+3\right)}\)=\(\dfrac{101}{1540}\)

\(\dfrac{1}{3}\)(\(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)+\(\dfrac{3}{11.14}\)+...+\(\dfrac{3}{x\left(x+3\right)}\))=\(\dfrac{101}{1540}\)

\(\dfrac{1}{3}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{11}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+3}\))=\(\dfrac{101}{1540}\)

\(\dfrac{1}{3}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\))=\(\dfrac{101}{1540}\)

\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\)=\(\dfrac{101}{1540}\) : \(\dfrac{1}{3}\)

\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\)=\(\dfrac{303}{1540}\)

\(\dfrac{1}{x+3}\)=\(\dfrac{1}{5}\)-\(\dfrac{303}{1540}\)

\(\dfrac{1}{x+3}\)=\(\dfrac{1}{308}\)

<=>1(x+3)=308.1

<=>1(x+3)=308

<=> x+3=308:1

<=> x+3=308

<=> x=308-3

<=> x=305

b,1+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{1}{x\left(x+1\right):2}\)=1\(\dfrac{1991}{1993}\)

\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{3984}{1993}\)\(2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3984}{1993}\)

\(2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3984}{1993}\)

\(2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3984}{1993}\)

\(1-\dfrac{1}{x+1}=\dfrac{3984}{1993}:2\)

\(1-\dfrac{1}{x+1}=\dfrac{1992}{1993}\)

\(\dfrac{1}{x+1}=1-\dfrac{1992}{1993}\)

\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)

<=>1(x+1)=1993.1

<=>1(x+1)=1993

<=> x+1=1993 : 1

<=> x+1=1993

<=> x=1993-1

<=> x=1992