\(1+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac...">
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8 tháng 7 2020

\(A=1+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}\)

\(< =>A=1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)

\(< =>2A=2+\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)

\(< =>2A=2+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}\)

\(< =>2A=\frac{5}{2}-\frac{1}{12}=\frac{29}{12}\)

\(< =>A=\frac{29}{12}.\frac{1}{2}=\frac{29}{24}\)

26 tháng 4 2019

A=1/8+1/24+1/48+1/80+1/120+1/168+1/224=>2A=2/8+2/24+2/48+2/80+2/120+2/168+2/224

2A=2/2*4+2/4*6+2/6*8+2/8*10+2/10*12+2/12*14+2/14*16

2A=1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12+1/12-1/14+1/14-1/16

2A=1/2-1/16

2A=7/16

A=7/16:2

A=7/32

10 tháng 5 2019

Gợi ý: đáp án \(\frac{15}{10}\)

10 tháng 5 2019

240240/240240+1/240240+240240/240240+1/240240+240240/240240+1/240240+240240/240240+1/240240+240240/240240+1/240240

=15/10

28 tháng 2 2019

giúp mk nhanh nhé

ai nhanh mk tk cho

28 tháng 2 2019

B1

a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)

\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)

\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)

\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)

\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)

\(x-\frac{11}{6}=1:\frac{3}{50}\)

\(x-\frac{11}{6}=\frac{50}{3}\)

\(x=\frac{50}{3}+\frac{11}{6}\)

\(x=\frac{37}{2}\)

b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)

\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)

\(\frac{5}{7}:x=-\frac{4}{15}\)

\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)

\(x=-\frac{75}{28}\)

c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)

\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)

\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)

\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)

\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)

\(\frac{2}{5}.x=-1\)

\(x=-1:\frac{2}{5}\)

\(x=-\frac{5}{2}\)

B2

a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)

\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)

\(=\frac{7}{6}.24:5-\frac{33}{10}\)

\(=28:5-\frac{33}{10}\)

\(=\frac{28}{5}-\frac{33}{10}\)

\(=\frac{56}{10}-\frac{33}{10}\)

\(=\frac{23}{10}\)

b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)

\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)

\(=\frac{61}{70}+0:\frac{25}{144}\)

\(=\frac{61}{70}+0\)

\(=\frac{61}{70}\)

19 tháng 4 2020

các bạn giúp mik với ạ

a) \(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)

\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)

\(A=2\cdot\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{15\cdot16}\right)\)

\(A=2\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(A=2\cdot\left(\frac{1}{4}-\frac{1}{16}\right)=2\cdot\frac{3}{16}=\frac{3}{8}\)

b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(B=\frac{5}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{25\cdot28}\right)\)

\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)

5 tháng 5 2020

Bài 1:

gọi số đó là x

ta có : \(\frac{-1}{12}< x< \frac{-1}{2}\)

hay :

\(\frac{-1}{12}< x< \frac{-6}{12}\)

vậy \(x\in\left\{\frac{-2}{12};\frac{-3}{12};\frac{-4}{12};\frac{-5}{12}\right\}\)

Tính tổng tất cả các phân số có mẫu số là 12  là :

\(\frac{-2}{12}+\frac{-3}{12}+\frac{-4}{12}+\frac{-5}{12}=\frac{-14}{12}=\frac{-7}{6}\)

bài 2:

\(A=1+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}\)

\(A=1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)

\(2A=2+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}\)

\(2A=3-\frac{1}{12}\)

\(A=\left(\frac{35}{12}\right):2=\frac{35}{24}\)