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20 tháng 4 2018

=\(\frac{3}{4}-\frac{5}{2}-\frac{5}{3}+\left(\frac{-1}{8}\right)=\frac{18}{24}-\frac{60}{24}-\frac{40}{24}-\frac{3}{24}=\frac{-85}{24}\)

22 tháng 4 2018

\(\frac{3}{4}-\left(\frac{5}{2}+\frac{5}{3}\right)+\left(\frac{-1}{2}\right)^3\)

\(=\frac{3}{4}-\frac{25}{6}+\frac{-1}{8}\)

\(=\frac{18-100-3}{24}\)

\(=\frac{-85}{24}\)

Tk mk nha!

24 tháng 5 2017

(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)

=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10

=8/15.48/63.120/143.3/8.35/48.9/10

=384/945.360/1144.315/480

=138240/1081080.315/480

=43545600/518918400=84/1001

24 tháng 5 2017

khó quá

21 tháng 3 2016

\(3\frac{1}{2}+4\frac{2}{5}=\left(3+4\right)+\left(\frac{1}{2}+\frac{2}{5}\right)=7+\frac{9}{10}=7\frac{9}{10}\)

nha....................................................

5 tháng 8 2016

\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)

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

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

\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)

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

\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)

\(=-2+3-\frac{31}{20}\)

\(=1-\frac{31}{20}=\frac{-11}{20}\)

6 tháng 8 2016

\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)

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

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

\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)

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

\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)

\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)

\(=1-\frac{31}{20}=\frac{-11}{20}\)

3/4-[(-5/3)-(1/12+2/9)]

3/4-[(-5/3)-11/36]

3/4+71/36

49/18

4 tháng 8 2020

\(\frac{3}{4}-\left[\left(-\frac{5}{3}\right)-\left(\frac{1}{12}+\frac{2}{9}\right)\right]\)

\(=\frac{3}{4}-\left[-\frac{5}{3}-\frac{11}{36}\right]\)

\(=\frac{3}{4}-\left(-\frac{71}{36}\right)=\frac{3}{4}+\frac{71}{36}\)

\(=\frac{49}{18}\)

25 tháng 6 2019

\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)

\(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)

\(\left(\frac{3}{4}+\frac{11}{4}+\frac{1}{2}\right)\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3\)

\(8+2+1-3\)

\(8\)

25 tháng 6 2019

#)Giải :

\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)

\(=\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)

\(=\left(\frac{3}{4}+\frac{11}{4}\right)+\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3+\frac{1}{2}\)

\(=\frac{7}{2}+2+1-3+\frac{1}{2}\)

\(=\frac{7}{2}+\frac{1}{2}\)

\(=4\)

24 tháng 10 2021

qwertyuiopasdfgggggghjkllzxcvbnmm,.//234567890-=`

24 tháng 10 2021
Chịu khó đọc lại đi dễ mà

x-[17/2-6/35]=-1/3

x-583/70=-1/3

x=-1/3+583/70

x=1679/210

vậy x=1769/210

[2/3-(x-7/4)]=9/2+5/4

[2/3-(x-7/4)]=23/4

(x-7/4)=23/4+2/3

(x-7/4)=77/12

x=77/12+7/4

x=49/6

vậy x=49/6

12 tháng 7 2016

Ta có: \(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+...+\frac{2}{2015^2}\right)< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)

\(=2\left(2+1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)=2\left(3-\frac{1}{2015}\right)=6-\frac{2}{2015}\)

Vậy D < 6.

12 tháng 7 2016

\(D=\frac{\left(2!\right)^2}{1^2}+\frac{\left(2!\right)^2}{3^2}+\frac{\left(2!\right)^2}{5^2}+\frac{\left(2!\right)^2}{7^2}+...+\frac{\left(2!\right)^2}{2015^2}\)

=>\(D=\frac{\left(1.2\right)^2}{1^2}+\frac{\left(1.2\right)^2}{3^2}+\frac{\left(1.2\right)^2}{5^2}+\frac{\left(1.2\right)^2}{7^2}+...+\frac{\left(1.2\right)^2}{2015^2}\)

=>\(D=\frac{2^2}{1^2}+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2015^2}\)

=>\(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}\right)\)

Ta có: \(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}< 2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)

=>\(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}\right)< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

Mà \(2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)\(=2\left(2+\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=2\left(2+1-\frac{1}{2015}\right)=2\left(3-\frac{1}{2015}\right)=6-\frac{6}{2016}< 6\)

=>\(D< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)< 6\)

=>D<6