Câu hỏi của Kang Daniel

Question

Tìm x biết$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}$

kudo shinichi · Accepted Answer

$\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}$$\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{2015}{2017}$$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}$$\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}$$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}$$\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}$$\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}$$\frac{1}{x+1}=\frac{1}{2017}$$\Rightarrow$x+1=2017$\Rightarrow$x=2017-1        x=2016Vậy x=2016Chúc bạn học tốt+-*/Câu trả lời: $\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}$$\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{2015}{2017}$$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}$$\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}$$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}$$\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}$$\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}$$\frac{1}{x+1}=\frac{1}{2017}$$\Rightarrow$x+1=2017$\Rightarrow$x=2017-1        x=2016Vậy x=2016Chúc bạn học tốt+-*/

Nguyễn Đình Toàn · Answer

x = -2018 nha bạn Câu trả lời: x = -2018 nha bạn

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