Câu hỏi của Vân Anh Đinh

Question

1 + 1 +  1  +.......+   1    = 101​ 5.8 8.4 11.14        x.(n+3) 1540 Dấu chấm là dấu nhân nha!!

Lung Thị Linh · Accepted Answer

$\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{n\left(n+3\right)}=\frac{101}{1540}$$\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{n\left(n+3\right)}=\frac{101}{1540}.3$$\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{303}{1540}$$\frac{1}{5}-\frac{1}{n+3}=\frac{303}{1540}$$\frac{1}{n+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$$\Rightarrow n=308-3=305$Câu trả lời: $\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{n\left(n+3\right)}=\frac{101}{1540}$$\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{n\left(n+3\right)}=\frac{101}{1540}.3$$\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{303}{1540}$$\frac{1}{5}-\frac{1}{n+3}=\frac{303}{1540}$$\frac{1}{n+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$$\Rightarrow n=308-3=305$

Hoàng Tử của dải Ngân Hà · Answer

$\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(n+3\right)}=\frac{101}{1540}$$\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}\right)+\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}\right)+\frac{1}{3}.\left(\frac{1}{11}-\frac{1}{14}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{n+3}=\frac{101}{1540}\right)$$\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{n}-\frac{1}{n+3}\right)=\frac{101}{1540}$$\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{n+3}\right)=\frac{101}{1540}$$\frac{1}{5}-\frac{1}{n+3}=\frac{303}{1540}$$\Rightarrow\frac{1}{n+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$$\Rightarrow n=308-3=305$kết quả việt đúng nhưng bước bị sai và thiếu,bạn phải làm chi tiết như mình mới đúngCâu trả lời: $\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(n+3\right)}=\frac{101}{1540}$$\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}\right)+\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}\right)+\frac{1}{3}.\left(\frac{1}{11}-\frac{1}{14}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{n+3}=\frac{101}{1540}\right)$$\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{n}-\frac{1}{n+3}\right)=\frac{101}{1540}$$\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{n+3}\right)=\frac{101}{1540}$$\frac{1}{5}-\frac{1}{n+3}=\frac{303}{1540}$$\Rightarrow\frac{1}{n+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$$\Rightarrow n=308-3=305$kết quả việt đúng nhưng bước bị sai và thiếu,bạn phải làm chi tiết như mình mới đúng

Edogawa Conan · Answer

$\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{n\left(n+3\right)}=\frac{101}{1540}$$\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{n\left(n+3\right)}=\frac{101}{1540}.3$$\frac{1}{5}-\frac{1}{n+3}=\frac{303}{1540}$$\frac{1}{n+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$$\Rightarrow n=308-3=305$Câu trả lời: $\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{n\left(n+3\right)}=\frac{101}{1540}$$\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{n\left(n+3\right)}=\frac{101}{1540}.3$$\frac{1}{5}-\frac{1}{n+3}=\frac{303}{1540}$$\frac{1}{n+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$$\Rightarrow n=308-3=305$

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