K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-1998}{1999}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{1998}{1999}=\dfrac{1}{1999}\)

29 tháng 8 2023

Có thể giải thik rõ hơn chút cho e đc ko ạ?

tính a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\) b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\) c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\) d)...
Đọc tiếp

tính

a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)

b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)

c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)

e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)

f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)

h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)

3
7 tháng 10 2017

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

7 tháng 10 2017

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

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

\(=3-1+\dfrac{1}{4}:2\)

\(=3-1+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

20 tháng 7 2018

A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)

= \(\left(-2\right).\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{215}{214}\right)\)

= \(\dfrac{\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right)...\left(-215\right)}{1.2.3.4...214}\)

= \(\dfrac{2.3.4.5...215}{1.2.3.4...214}\)

= \(\dfrac{215}{1}=215\)

20 tháng 7 2018

B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)....\left(-1\dfrac{1}{299}\right)\)

= \(\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{300}{299}\right)\)

= \(\dfrac{\left(-3\right).\left(-4\right).\left(-5\right)...\left(-300\right)}{2.3.4...299}\)

= \(\dfrac{3.4.5...300}{2.3.4.5...299}\)

= \(\dfrac{300}{2}=150\)

\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right)\left(\dfrac{1}{2016}-1\right)\left(\dfrac{1}{2017}-1\right)\\ A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)...\left(-\dfrac{2014}{2015}\right)\left(-\dfrac{2015}{2016}\right)\left(-\dfrac{2016}{2017}\right)\\ A=\dfrac{1.2.3.4...2014.2015.2016}{2.3.4...2015.2016.2017}=\dfrac{1}{2017}\)

\(B=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right)\left(-1\dfrac{1}{2016}\right)\left(-1\dfrac{1}{2017}\right)\\ B=\left(-\dfrac{3}{2}\right)\left(-\dfrac{4}{3}\right)\left(-\dfrac{5}{4}\right)...\left(-\dfrac{2016}{2015}\right)\left(-\dfrac{2017}{2016}\right)\left(-\dfrac{2018}{2017}\right)\\ B=\dfrac{3.4.5...2016.2017.2018}{2.3.4...2015.2016.2017}=\dfrac{2018}{2}=1009\)

\(M=A.B=\dfrac{1}{2017}.1009=\dfrac{1009}{2017}\)

29 tháng 8 2023

\(...A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)....\left(-\dfrac{1998}{1999}\right).\)

Số dấu trừ là : \(\left(1998-1\right):1+1=1998\) là số chẵn

\(\Rightarrow A=\dfrac{1.2.3...1998}{2.3.4...1999}\)

\(\Rightarrow A=\dfrac{1}{1999}\)

29 tháng 8 2023

gợi ý nè

tính hết mấy cái hiệu trong ngoặc rồi nhân lại

vì kết thúc ở số 1999

nên sẽ có 1999 dấu -

nên kq là âm

nhân ra rồi triệt tiêu đi

\(A=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-2016}{2017}=\dfrac{1}{2017}\)

\(B=\dfrac{-3}{2}\cdot\dfrac{-4}{3}\cdot...\cdot\dfrac{-2018}{2017}=\dfrac{2018}{2}=1009\)

\(A\cdot B=\dfrac{1009}{2017}\)

8 tháng 10 2017

\(a.\)

\(\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)

\(=\left[6.\dfrac{1}{9}+1+1\right]:\left(-\dfrac{4}{3}\right)\)

\(=\left(\dfrac{8}{3}\right):\left(-\dfrac{4}{3}\right)\)

\(=\left(\dfrac{8}{3}\right).\left(-\dfrac{3}{4}\right)\)

\(=-2\)

\(b.\)

\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)

\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\)

\(=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}\)

\(=\dfrac{72}{5}\)

19 tháng 4 2017

Thực hiện các phép tính:

a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14

b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;

c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)

d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113.

Hướng dẫn làm bài:

a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14

=9,6.52−(250−1712)×4=9,6.52−(250−1712)×4

=4,8.5−(1000−173)=4,8.5−(1000−173)

=24−1000+173=24−1000+173

=−976+173=−976+173

=−97013=−97013

b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;

=518−1,456×257+92.45=518−1,456×257+92.45

=518−0,208×25+185=518−0,208×25+185

=518−5,2+185=518−5,2+185

=25−468+32490=25−468+32490

=−11990=−11990

c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)

=(12+45−43).(2310+10725−3225)=(12+45−43).(2310+10725−3225)

=(15+24−4030).(2310+10725−3225)=(15+24−4030).(2310+10725−3225)

=(15+24−4030).(115+214−6450)=(15+24−4030).(115+214−6450)

=−130.26550=−130.26550

=−53300=−53300

d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113

=−60:[14+12×(−12)]+1.13=−60:[14+12×(−12)]+1.13

=−60:[−14−14]+113=−60:[−14−14]+113

=−60:(12)+113=−60:(12)+113

=120+113=120+113

=12113

19 tháng 4 2017

a) \(9,6.2\dfrac{1}{2}-\left(2.125-1\dfrac{5}{12}\right):\dfrac{1}{4}\)

\(=9,6.\dfrac{5}{2}-\left(250-\dfrac{17}{12}\right).4\)

\(=4,8.5-\left(1000-\dfrac{17}{3}\right)\)

\(=24-1000+\dfrac{17}{3}\)

\(=-976+\dfrac{17}{3}=-970\dfrac{1}{3}\)

b) \(\dfrac{5}{18}-1,456:\dfrac{7}{25}+4,5.\dfrac{4}{5}\)

\(=\dfrac{5}{18}-1,456.\dfrac{25}{7}+\dfrac{9}{2}.\dfrac{4}{5}\)

\(=\dfrac{5}{18}-0,208.25+\dfrac{18}{5}\)

\(=\dfrac{5}{18}-5,2+\dfrac{18}{5}\)

\(=-\dfrac{119}{90}\)

c) \(\left(\dfrac{1}{2}+0,8-1\dfrac{1}{3}\right).\left(2,3+4\dfrac{7}{25}-1,28\right)\)

\(=\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{4}{3}\right).\left(\dfrac{23}{10}+\dfrac{107}{25}-\dfrac{32}{25}\right)\)

\(=-\dfrac{1}{30}.\dfrac{265}{50}=-\dfrac{53}{300}\)

d) \(\left(-5\right).12:\left[\left(-\dfrac{1}{4}\right)+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{1}{3}\)

\(=-60:\left[\dfrac{1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right]+1.\dfrac{1}{3}\)

\(=-60:\left[-\dfrac{1}{4}-\dfrac{1}{4}\right]+1\dfrac{1}{3}\)

\(=-60:\left(\dfrac{1}{2}\right)+1\dfrac{1}{3}\)

\(=121\dfrac{1}{3}\)

30 tháng 11 2017

Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi

30 tháng 11 2017

0 cần trả lời hết cũng đc