\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) +
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8 tháng 4 2017

Ta có:

\(2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)

\(=2.1+2.\dfrac{1}{2}+2.\dfrac{1}{4}+2.\dfrac{1}{8}+...+2.\dfrac{1}{1024}\)

\(=2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)

Ta lại có:

\(A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)

\(=2-\dfrac{1}{1024}=\dfrac{2047}{1024}\)

8 tháng 4 2017

giỏi ghê ta

27 tháng 7 2017

Các bạn ơi,mình ghi thiếu,còn 3 câu nữa nha!!!~~nya

e)| \(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\) |-(-22).\(\dfrac{1}{3}\)(0,75-\(\dfrac{1}{7}\))=\(\dfrac{-5}{13}\):2\(\dfrac{9}{13}\)-0,5.(\(\dfrac{-2}{3}\))

f)| 5x+21 | = | 2x -63 |

g) -45 - |-3x-96 | - 54=-207

Làm ơn giúp mình với ạ!Mình đang cần gấp lắm trong ngày hôm nay ạ!!!Mình xin cảm ơn các bạn nhiều nhiều lắm luôn đó!!!Thank you very much!!!(^-^)

1 tháng 8 2017

a, (\(\dfrac{2}{9}\)(6x - \(\dfrac{3}{4}\)) - 3(\(\dfrac{1}{4}x-\dfrac{1}{5}\)) = \(\dfrac{-8}{15}\)

<=> (\(\dfrac{4}{3}x-\dfrac{1}{6}\)) - (\(\dfrac{3}{4}x-\dfrac{3}{5}\)) = \(\dfrac{-8}{15}\)

<=> \(\dfrac{4}{3}x-\dfrac{1}{6}-\dfrac{3}{4}x+\dfrac{3}{5}=\dfrac{-8}{15}\)

<=> \(\dfrac{7}{12}x+\dfrac{13}{30}=\dfrac{-8}{15}\)

<=> \(\dfrac{7}{12}x=\dfrac{-8}{15}-\dfrac{13}{30}\)

<=> \(\dfrac{7}{12}x=-\dfrac{29}{30}\)

<=> x = \(-\dfrac{58}{35}\)
@Nguyễn Gia Hân

3 tháng 4 2017

B=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

B=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

B=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

B= 1-\(\dfrac{1}{8}\)

B= \(\dfrac{7}{8}\)

24 tháng 4 2017

\(A=\dfrac{5}{9}-\dfrac{5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\ =\dfrac{5}{9}+\dfrac{-5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\= \left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{-5}{8}+\dfrac{-3}{8}\right)\\ =1+1+\left(-1\right)\\ =2+\left(-1\right)\\ =1\)

31 tháng 3 2017

a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)

b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)

c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)

d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)

lưu ý mk ko chép đầu bài

31 tháng 3 2017

mình cần gấp lắm đến chiều mai là phải nộp rùi

giúp mình nha thanks cá bạn trước vuiko có tâm trạng mà cười nữalolanglimdim

12 tháng 3 2017

Bài 2:

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)

Vậy A < 2

Bài 3:

D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)

\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)

Bài 4:

A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)

\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)

\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

20 tháng 3 2017

a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)

=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)

=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)

=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)

b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)

=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)

=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)

=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)

c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)

=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)

= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)

d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)

=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)

=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)

e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)

=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)

21 tháng 3 2017

2) Tinh nhanh:

a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)

= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)

= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)

= \(\dfrac{5}{598}\)

21 tháng 3 2017

b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)

= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)

22 tháng 3 2017

\(a)\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{13}{4}-\dfrac{7}{-24}\)

\(=\dfrac{85}{24}\)

\(b)\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{-3}{56}-\dfrac{3}{28}\)

\(=\dfrac{-9}{56}\)

\(c)\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)

\(=\dfrac{13}{12}\)\(+\dfrac{-2}{3}\)

\(=\dfrac{5}{12}\)

\(d)\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-1}{14}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-23}{126}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-4}{7}+\dfrac{4}{7}\)

\(=0\)

\(e)\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{-5}{56}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{83}{56}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{305}{168}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{47}{24}+\dfrac{5}{-8}\)

\(=\dfrac{4}{3}\)

22 tháng 3 2017

Bài 2 : Tính

a) \(\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{18}{24}-\dfrac{-60}{24}-\dfrac{-4}{24}\)

\(=\dfrac{18-\left(-60\right)-\left(-7\right)}{24}\)

\(=\dfrac{85}{24}\)

b) \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{32}{56}+\dfrac{-35}{56}-\dfrac{6}{56}\)

\(=\dfrac{32+\left(-35\right)-6}{56}\)

\(=\dfrac{-9}{56}\)

c) \(\dfrac{7}{36}-\dfrac{8}{9}+\dfrac{-2}{3}\)

\(=\dfrac{7}{36}-\dfrac{32}{36}+\dfrac{-24}{36}\)

\(=\dfrac{7-32+\left(-24\right)}{36}\)

\(=\dfrac{-49}{36}\)

d) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-9}{18}+\dfrac{3}{7}-\dfrac{2}{18}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\left(\dfrac{-9}{18}+\dfrac{-7}{18}-\dfrac{2}{18}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=\left(-1\right)+1\)

\(=0\)

e) \(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{11}{7}\right)+\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\dfrac{1}{3}\)

\(=2+\left(-1\right)+\dfrac{1}{3}\)

\(=1+\dfrac{1}{3}\)

\(=\dfrac{4}{3}\)

12 tháng 3 2017

banhA = \(\dfrac{17}{15}.\dfrac{-31}{125}.\dfrac{1}{2}.\dfrac{10}{17}.\dfrac{-1}{8}\)

= \(\dfrac{17.\left(-31\right).1.10.\left(-1\right)}{15.125.2.17.8}\)

= \(\dfrac{17.\left[\left(-31\right).\left(-1\right)\right].1.2.5}{5.3.125.17.4.2}\)

= \(\dfrac{31.1}{3.125.4}\)

= \(\dfrac{31}{1500}\)

banhB = \(\left(\dfrac{11}{4}.\dfrac{-5}{9}-\dfrac{4}{9}.\dfrac{11}{4}\right).\dfrac{8}{33}\)

= \(\left[\dfrac{11}{4}.\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)\right].\dfrac{8}{33}\)

= \(\left(\dfrac{11}{4}.\dfrac{-9}{9}\right).\dfrac{8}{33}\)

= \(\left[\dfrac{11}{4}.\left(-1\right)\right].\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)

= \(\dfrac{-11}{4}.\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)

= \(\dfrac{\left(-11\right).4.2}{4.\left(-11\right)\left(-3\right)}\)

= \(\dfrac{2}{-3}\)

Ok nhá!