giai gium

T

thaonguyenkmhd

phần a/

$ \frac{1}{5.8} + \frac{1}{8.11} + \frac{1}{11.14} + ... + \frac{1}{x( x+3 )} = \frac{101}{1540} $

\Leftrightarrow$\frac{1}{3} ( \frac{3}{5.8} + \frac{3}{8.11} + \frac{3}{11.14} + ... + \frac{3}{x( x+3 )} ) = \frac{101}{1540}$

\Leftrightarrow$\frac{1}{3} ( \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + \frac{1}{11} - \frac{1}{14} ... + \frac{1}{x} - \frac{1}{x + 3} ) = \frac{101}{1540}$

\Leftrightarrow$ \frac{1}{3} ( \frac{1}{5} - \frac{1}{x + 3} ) = \frac{101}{1540}$

\Leftrightarrow$ \frac{1}{5} - \frac{1}{x + 3} = \frac{101}{1540} : \frac{1}{3} = \frac{303}{1540}$

\Leftrightarrow$ \frac{1}{x + 3} = \frac{1}{5} - \frac{303}{1540} $

\Leftrightarrow$ \frac{1}{x + 3} = \frac{1}{308}$

\Leftrightarrow$ x + 3 = 308$

\Leftrightarrow$ x = 305$

Vậy $x = 305 $
 
H

hiensau99

$1+ \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + ... + \frac{1}{x( x+2 ):2} = \frac{1991}{1993} $

$\Longrightarrow \frac{1}{2}+\frac{1}{6} + \frac{1}{12} + \frac{1}{20} + ... + \frac{1}{x( x+2 )} = \frac{1991}{1993.2}$

$\Longrightarrow \frac{1}{1.2}+\frac{1}{2.3} + \frac{1}{3.4} + \frac{1}{4.5} + ... + \frac{1}{x.( x+2 )} = \frac{1991}{1993.2}$

$\Longrightarrow 1-\frac{1}{2}+\frac{1}{2} - \frac{1}{3} + \frac{1}{4} -\frac{1}{5} + ... + \frac{1}{x}-\frac{1}{x+2} = \frac{1991}{1993.2}$

$\Longrightarrow 1-\frac{1}{x+2} = \frac{1991}{3986}$

$\Longrightarrow \frac{1}{x+2} = \frac{1991}{3986}-1$

$\Longrightarrow \frac{1}{x+2} = \frac{-1995}{3986}$

$\Longrightarrow x+2 = \frac{-3986}{1995}$

$\Longrightarrow x = \frac{-7976}{1995}$


 
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