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

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

\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)

\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)

\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)

\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)

\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)

\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)

\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)

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

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

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

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

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

\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)

\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)

\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)

\(B=\frac{1}{2}.\frac{5}{24}\)

\(B=\frac{5}{48}\)

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

\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)=\frac{1}{3}.\frac{18}{19}=\frac{6}{19}\) VẬY: \(A=\frac{6}{19}\)

\(B=\frac{1}{4.2.4}+\frac{1}{4.4.6}+...+\)\(\frac{1}{4.10.12}\)

\(B=\frac{1}{4}\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{10.12}\right)\)\(\Rightarrow B=\frac{1}{8}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{10}-\frac{1}{12}\right)\)

                                                                               \(\Rightarrow B=\frac{1}{8}\left(\frac{1}{2}-\frac{1}{120}\right)=\frac{1}{8}.\frac{59}{120}=\frac{59}{960}\)

đây là toán lớp 5 cơ mà

a)A=\(\frac{1}{1x4}\)+\(\frac{1}{4x7}\)+...+\(\frac{1}{16x19}\)

A=\(\frac{1}{3}\)x3x(\(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+.......+\(\frac{1}{16.19}\)

A=\(\frac{1}{3}\)x(\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+............+\(\frac{3}{16.19}\))

A=\(\frac{1}{3}\)x(1-1/4+1/4-1/7+......+1/13-1/16+1/16-1/19)

A=\(\frac{1}{3}\)x(1-\(\frac{1}{19}\))

A=\(\frac{1}{3}\)x\(\frac{18}{19}\)

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

câu b tương tự tách mẫu ra thôi

mik ko hiểu cách làm của bạn cho lắm ,có thể làm rõ hơn đc ko ???

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3x+2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{2}-\frac{1}{3x+5}=\frac{12}{25}\)

\(\frac{1}{3x+5}=\frac{1}{50}\)

=> 3x+5 = 50

3x = 45

x = 15

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)

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

Ta thấy :

 \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)

\(.........\)

\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)

\(\Rightarrow A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\)

\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)

đáp án = \(\frac{297}{100}\)

đúng không?

kết bạn với mh nha

dm thằng chó

Nguyen Trong Duc đang chửi ai vậy ?

các bạn có thể giải nhanh một chút đc ko

\(x=-\frac{5}{3}\)

\(\frac{3x+5}{2}+\frac{3x+5}{4}+\frac{3x+5}{6}=\frac{3x+5}{8}\)

\(\frac{3x+5}{2}+\frac{3x+5}{4}+\frac{3x+5}{6}-\frac{3x+5}{8}=0\)

\(\left(3x-5\right)\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}-\frac{1}{8}\right)=0\)

\(\Rightarrow x=-\frac{5}{3}\)

Một cách # dài dòng hơn.

\(\frac{3x+5}{2}+\frac{3x+5}{4}+\frac{3x+5}{6}=\frac{3x+5}{8}\)

\(\Leftrightarrow\frac{3}{2}x+\frac{5}{2}+\frac{3}{4}x+\frac{5}{4}+\frac{3}{6}x+\frac{5}{6}=\frac{3}{8}x+\frac{5}{8}\)

\(\Leftrightarrow\left(\frac{3}{2}x+\frac{3}{4}x+\frac{3}{6}x\right)+\left(\frac{5}{2}+\frac{5}{4}+\frac{5}{6}\right)=\frac{3}{8}x+\frac{5}{8}\)

\(\Leftrightarrow\frac{11}{4}x+\frac{55}{12}=\frac{3}{8}x+\frac{5}{8}\)

\(\Leftrightarrow\frac{11}{4}x+\frac{55}{12}-\frac{3}{8}x=\frac{5}{8}\)

\(\Leftrightarrow\frac{19}{8}x+\frac{55}{12}=\frac{5}{8}\)

\(\Leftrightarrow\frac{19}{8}x=\frac{5}{8}-\frac{55}{12}\)

\(\Leftrightarrow\frac{19}{8}x=-\frac{95}{24}\)

\(\Leftrightarrow x=-\frac{95}{24}:\frac{19}{8}=-\frac{5}{3}\)

\(\Rightarrow x=-\frac{5}{3}\)