Các bạn giúp tớ với, sáng mai mình học rồi.
a)\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
a) \(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+...+\(\dfrac{1}{x\left(x+3\right)}\)=\(\dfrac{101}{1540}\)
=\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
=\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}=\dfrac{1}{308}\)
=> x+3=308
=> x=308-3
=> x=305
b) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{1991}{1993}\)
= \(\dfrac{1}{2}.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)
= \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{3984}{3986}\)
= \(2-\dfrac{1}{1.2}+3-\dfrac{2}{2.3}+4-\dfrac{3}{3.4}+...+x+1-\dfrac{x}{x\left(x+1\right)}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
= \(1-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
=> \(\dfrac{1}{x+1}=1-\dfrac{3984}{3986}\)
=> \(\dfrac{1}{x+1}=\dfrac{2}{3986}=\dfrac{1}{1993}\)
=> x+1= 1993
=> x= 1993-1
=> x=1992
