Câu hỏi của nguyễn đức anh

Question

$\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{10}$+...+$\frac{2}{x\left(x+1\right)}$=$\frac{2011}{2013}$

Đức Phạm · Accepted Answer

$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}$$\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}$$\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}$$\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}$$\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}$$\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}\div2$$\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}$$\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}$$\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}$$\Leftrightarrow x=2013-1=2012$Câu trả lời: $\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}$$\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}$$\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}$$\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}$$\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}$$\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}\div2$$\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}$$\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}$$\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}$$\Leftrightarrow x=2013-1=2012$

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