toán khó lớp 8 đây

H

hien_vuthithanh

$\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+ \dfrac{2}{x.(x+1)}=1\dfrac{2013}{2015}$
\Leftrightarrow $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{2}{3.4}+...+ \dfrac{1}{x.(x+1)}=\dfrac{2014}{2015}$
\Leftrightarrow $1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2014}{2015}$
\Leftrightarrow $1-\dfrac{1}{x+1}=\dfrac{2014}{2015}$
\Leftrightarrow $\dfrac{1}{x+1}=\dfrac{1}{2015}$
\Rightarrow Giải tìm x
 
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