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`3x-15/(5*8)-15/(8*11)-15/(11*14)-...-15/(47*50)=2 1/10`
`3x-(15/(5*8)+15/(8*11)+15/(11*14)+...+15/(47*50))=21/10`
`3x-5(3/(5*8)+3/(8*11)+3/(11*14)+...+3/(47*50))=21/10`
`3x-5(1/5-1/8+1/8-1/11+1/11-1/14+...+1/47-1/50)=21/10`
`3x-5(1/5-1/50)=21/10`
`3x-5*9/50=21/10`
`3x-9/10=21/10`
`3x=21/10+9/10`
`3x=3`
`x=1`
<=> \(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
<=>\(x+3=308\)
<=>\(x=305\)
a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)
b.
\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)
\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)
\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)
\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)
đặt \(\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{65.68}\right)\)là A
Ax=\(\frac{19}{68}+\frac{7}{34}=\frac{33}{68}\)
3A=\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{11}{8.11}+...+\frac{1}{65.68}\right)\)
3A=\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\)
3A=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{65}-\frac{1}{68}\)
3A=\(\frac{1}{2}-\frac{1}{68}=\frac{33}{68}\)
A=33/68:3=11/68
\(\Rightarrow\)33/68:11/68=3
vậy x= 3
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(=3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x.3}=\frac{303}{1540}\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{308}\)
\(x+3=308\)
\(\Rightarrow x=305\)