Bài 3: Tìm x:
a) \(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+......+\(\dfrac{1}{x\left(x+3\right)}\)= \(\dfrac{101}{1540}\)
b) 1+ \(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+......+ \(\dfrac{1}{x\left(x+1\right):2}\)=1\(\dfrac{1991}{1993}\)
Ai biết làm bài này thì giúp mik nhé!
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\(a,\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.3.\left[\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left(\dfrac{1}{5-1}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}.\dfrac{1}{3}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{3}-\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
\(b,1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
\(\dfrac{1}{2}.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x.\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{8}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+x+1-\dfrac{x}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=1-\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)
=>\(x+1=1993\)
\(x=1993-1\)
\(x=1992\)
