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3 tháng 5 2020

\(125\%\cdot\left(-\frac{1}{2}\right)^2\div\left(1\frac{5}{6}-1,5\right)+2016^0\)

\(=\frac{125}{100}\cdot\frac{1}{4}\div\left(\frac{11}{6}-\frac{3}{2}\right)+1\)

\(=\frac{5}{4}\cdot\frac{1}{4}\div\left(\frac{11}{6}-\frac{9}{6}\right)+1\)

\(=\frac{5}{16}\div\frac{2}{6}+1\)

\(=\frac{15}{16}+1=\frac{31}{16}\)

3 tháng 5 2020

         \(125\%.\left(\frac{-1}{2}\right)^2:\left(1\frac{5}{6}-1,5\right)+2016^0\)

\(=\frac{125}{100}.\frac{1}{4}:\left(\frac{11}{6}-\frac{3}{2}\right)+1\)

\(=\frac{5}{16}:\left(\frac{11}{6}-\frac{9}{6}\right)+1\)

\(=\frac{5}{16} :\frac{1}{3}+1\)

\(=\frac{5}{16} .\frac{3}{1}+1\)

\(=\frac{15}{16}+1\)

\(=\frac{31}{16}\)

9 tháng 6 2017

Theo đề ta có:

\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)\)\(-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)

= \(5+\)\(\frac{1}{5}-\frac{2}{9}\)-\(2+\frac{1}{23}+2+\frac{3}{5}+\frac{5}{6}-8+\frac{2}{3}-\frac{1}{18}\)

=\(\left(5+2-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{5}{6}-\frac{2}{3}+\frac{1}{18}\right)+\frac{1}{23}\)

=  -1 +\(\frac{4}{5}\)\(-\frac{-11}{9}\)+\(\frac{1}{23}\)

= -1 +\(\frac{4}{5}+\frac{11}{9}+\frac{1}{23}\)

10 tháng 6 2017

\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)

\(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+2+\frac{3}{5}-\frac{5}{6}-8+\frac{2}{3}+\frac{1}{18}\)

\(\left(5-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{1}{18}-\frac{2}{3}\right)-\left(2-2\right)+\frac{1}{23}-\frac{5}{6}\)

\(\left(-3\right)+\frac{4}{5}+\frac{1}{2}+\frac{1}{23}-\frac{5}{6}\)

\(\left(\left(-3\right)+\frac{4}{5}+\frac{1}{2}-\frac{5}{6}\right)+\frac{1}{23}\)

\(-\frac{38}{15}+\frac{1}{23}\)

\(-\frac{859}{345}\)

3 tháng 11 2018

trong tích trên có 1 thừa số như thế này:

\(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{125}\right)\)

=0

=> tích trên bằng 0

8 tháng 9 2017

a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)

\(=8+3.1+4:\frac{1}{2}\)

\(=8+3+8=19\)

b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)

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

\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)

d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)

25 tháng 4 2018

= [ 60/90 - ( 12/15 + 10/15) ] : 6/5

= ( 2/3 - 22/15 ) x 5/6

= ( 10/15 - 22/15 ) x 5/6

= -12/15 x 5/6

= -60/90

= -2/3

25 tháng 4 2018

( 15/10 x 4/9 - ( 4/5 + 2/3 )) : 6/5

( 15/10 x 4/9 - ( 12/15 + 10/15 )) : 6/5

(30/45 - 66/45 ) : 6/5

-12/15 : 6/5 ( đã rút gọn -36/45 = -12/15 )

-2/3

k mk na <3

23 tháng 6 2015

a)\(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}\cdot\frac{1}{3}=4+\frac{1}{27}=\frac{108}{27}+\frac{1}{27}=\frac{109}{27}\)

b)\(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[\left(-8\right)\cdot2\right]=8+3-16=-5\)

23 tháng 6 2015

a/ \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=5-1+\frac{1}{27}=4+\frac{1}{27}=\frac{109}{27}\)

b/ \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[-8:\frac{1}{2}\right]=11+-16=-5\)

28 tháng 2 2018

\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)

\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)

\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)

2.

\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)

\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)

3.

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)

\(=\frac{1}{100}\)

4.

\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)

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

28 tháng 2 2018

mình chỉ làm được câu 3 thôi

có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)

\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)

\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)

\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)

\(=\frac{-1}{100}\)

22 tháng 3 2018
  1.  (1/2-1/3-1/6).(3/8+34/88-345/888)​​​

= (3/6-2/6-1/6).(3/8+34/88-345/888)

= 0.(3/8+434/88-345/888)=0

      2.  8/3.2/5.3/8.10.19/92

= (8/3.3/8).(2/5.10).19/92

= 1.4.19/92

= 76/92

22 tháng 3 2018

1) \(\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\left(\frac{3}{8}+\frac{34}{88}+\frac{345}{888}\right)=\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\left(\frac{3}{8}+\frac{34}{88}+\frac{345}{888}\right)\)

                                                                                   \(=\left(\frac{1}{6}-\frac{1}{6}\right)\left(\frac{3}{8}+\frac{34}{88}+\frac{345}{888}\right)\)

                                                                                   \(=0\cdot\left(\frac{3}{8}+\frac{34}{88}+\frac{345}{888}\right)=0\)(số nào nhân với 0 cũng bằng 0)

2) \(\frac{8}{3}\cdot\frac{2}{5}\cdot\frac{3}{8}\cdot10\cdot\frac{19}{92}=\frac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\)

\(=\frac{2\cdot10\cdot19}{5\cdot92}=\frac{2\cdot2\cdot5\cdot19}{5\cdot2\cdot2\cdot23}=\frac{19}{23}\)