Câu hỏi của Nhân Văn

Question

$\frac{5}{1.4}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}

$

$\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}$

Huy Thắng Nguyễn · Accepted Answer

lúc đầu ý bn là 5/1.3 đúng k, mk chỉnh lại như thế cho tiện nhé

a) $\frac{5}{1\times3}+\frac{5}{3\times5}+\frac{5}{5\times7}+...+\frac{5}{99\times101}$

$=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)$

$=\frac{5}{2}\left(1-\frac{1}{101}\right)$

$=\frac{5}{2}\times\frac{100}{101}=\frac{250}{101}$

b) $\frac{3^2}{8\times11}+\frac{3^2}{11\times14}+\frac{3^2}{14\times17}+...+\frac{3^2}{197\times200}$

$=\frac{9}{8\times11}+\frac{9}{11\times14}+\frac{9}{14\times17}+...+\frac{9}{197\times200}$

$=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)$

$=3\left(\frac{1}{8}-\frac{1}{200}\right)$

$=3\times\frac{3}{25}=\frac{9}{25}$Câu trả lời: lúc đầu ý bn là 5/1.3 đúng k, mk chỉnh lại như thế cho tiện nhé

a) $\frac{5}{1\times3}+\frac{5}{3\times5}+\frac{5}{5\times7}+...+\frac{5}{99\times101}$

$=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)$

$=\frac{5}{2}\left(1-\frac{1}{101}\right)$

$=\frac{5}{2}\times\frac{100}{101}=\frac{250}{101}$

b) $\frac{3^2}{8\times11}+\frac{3^2}{11\times14}+\frac{3^2}{14\times17}+...+\frac{3^2}{197\times200}$

$=\frac{9}{8\times11}+\frac{9}{11\times14}+\frac{9}{14\times17}+...+\frac{9}{197\times200}$

$=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)$

$=3\left(\frac{1}{8}-\frac{1}{200}\right)$

$=3\times\frac{3}{25}=\frac{9}{25}$

Trần Quang Hưng · Answer

Ta có $\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}$

$\Rightarrow3^2.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)$

$\Rightarrow9.\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)$

$\Rightarrow3.\left(1-\frac{1}{200}\right)$

$\Rightarrow3.\frac{199}{200}=\frac{597}{200}$Câu trả lời: Ta có $\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}$

$\Rightarrow3^2.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)$

$\Rightarrow9.\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)$

$\Rightarrow3.\left(1-\frac{1}{200}\right)$

$\Rightarrow3.\frac{199}{200}=\frac{597}{200}$