\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+...+\frac{1}{4850}\)

\(=\frac{2}{4}+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+...+\frac{2}{9700}\)

\(=2.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)

\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

sao lại có 2/3 ở đây

Ta có :

\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)

Giá trị không đổi khi cả tử và mẫu cùng nhân với 2, ta được :

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)

\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{19}\right)=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)

\(A=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}=\frac{1}{2}.\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{19}\right)=\frac{9}{19}\)

Đặng Minh Triều làm sai. \(\frac{1}{2}.\frac{1}{14}\ne\frac{1}{28}\)

đề kiu tính hợp lý hay tính bình thường

12/19

12/19

N = 2/4+2/28+2/70+2/130+2/208+2/304

N = 2/1.4+2/4.7+2/7.10+2.......

c1: Quy đồng mẫu rồi cộng tử

c1:Phân tích mẫu rồi triệt tiêu

đáp án băng 12/19

Bằng 12/19

k nhé

khó quá

12/19 tk m nhé

\(C=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(C=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)

\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)

\(C=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)

\(C=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(C=\frac{2}{3}.\left(1-\frac{1}{19}\right)\)

\(C=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)

C= 1/2 +1/14 +1/35 +1/65 +1/104+1/152 =12/19

\(C=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(C=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\)

\(C=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)

\(C=\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{1}{3}.\left(\frac{1}{10}-\frac{1}{13}\right)+\frac{1}{3}.\left(\frac{1}{13}-\frac{1}{16}\right)+\frac{1}{3}.\left(\frac{1}{16}-\frac{1}{19}\right)\)

\(C=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(C=\frac{1}{3}.\left(1-\frac{1}{19}\right)\)

\(C=\frac{1}{3}.\frac{18}{19}\)

\(C=\frac{6}{19}\)

\(\frac{12}{19}\)

M=1/2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19

M=1/2-1/2+1/7-1/7+1/5-1/5+1/13-1/13+1/8-1/8+1/19

M=1/2-1/19

M=17/38

 Bằng 12/19

Ta có: \(\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+...+\frac{2}{x^2+3x}=\frac{1}{9}\)

\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}\cdot\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{2}{3}\cdot\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{2}{3}\cdot\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{9}\)

\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{9}\)

\(\Leftrightarrow\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{x+3}\right)=\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{4}-\frac{1}{x+3}=\frac{1}{9}:\frac{2}{3}=\frac{1}{9}\cdot\frac{3}{2}=\frac{1}{6}\)

\(\Leftrightarrow-\frac{1}{x+3}=\frac{1}{6}-\frac{1}{4}=\frac{2}{12}-\frac{3}{12}=-\frac{1}{12}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{12}\)

\(\Leftrightarrow x+3=12\)

hay x=9

Vậy: x=9

a có: 

hay x=9

Vậy: x=9