Ôn tập toán 8

DN

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

LF
4 tháng 1 2017 lúc 6:19

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

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{1991}{1993}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)

\(\Rightarrow x+1=1993\Rightarrow x=1992\)

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