K
Khách

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

8 tháng 8 2018

\(1\frac{13}{15}\cdot\left(0,5\right)^2\cdot3-\left(\frac{8}{15}+1\frac{19}{60}\right):1\frac{19}{60}\)

\(=\frac{28}{15}\cdot\frac{1}{4}\cdot\frac{3}{1}-\frac{37}{20}\cdot\frac{60}{79}\)

\(=\frac{7}{5}-\frac{111}{79}\)

\(=\frac{-2}{395}\)

1 tháng 4 2019

=28/15.1/4.3-8/15-79/60.60/79

=7.4/15.1/4.3-8/15-1

=7/15.3-8/15-1

=21/15-8/15-1

=13/15-1

=-2/15

4 tháng 10 2021

yutyugubhujyikiu

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)                                   

b)

\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)          

d)

\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ =  - 2\end{array}\)

18 tháng 12 2016

a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)

= \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\)

= \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\)

=\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\)

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

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

b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\)

= \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\)

= \(-12:\left\{\frac{-1}{12}\right\}^2\)

= \(-12:\frac{1}{144}\)

= \(-12.144\)

= -1728

c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)

= \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\)

= \(\frac{7}{23}.\left(\frac{-23}{6}\right)\)

= \(\frac{-7}{6}\)

d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)

= \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)

= \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\)

= \(10.\frac{7}{5}\)

= 14

 

e) (1+2314).(0,834)2

= (1+2314).(\(\frac{4}{5}\)34)2

= \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)

= \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)

= \(\frac{17}{20}.\frac{1}{400}\)

= \(\frac{17}{8000}\)

 

16 tháng 7 2016

a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)

\(=1+1+0,5\)

\(=2,5\)

b)\(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)

\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)

\(=\frac{3}{7}.\left(-14\right)\)

\(=-6\)

c)\(9.\left(-\frac{1}{3}\right)^3+\frac{1}{3}\)

\(=9.\left(-\frac{1}{3}\right)^3+9.\frac{1}{27}\)

\(=9.\left[\left(-\frac{1}{3}\right)^3+\frac{1}{27}\right]\)

\(=9.0=0\)

d)\(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)

\(=\left(15\frac{1}{4}-25\frac{1}{4}\right):\left(-\frac{5}{7}\right)\)

\(=\left(-10\right):\left(-\frac{5}{7}\right)\)

\(=14\)

 

30 tháng 11 2017

\(\frac{11}{15}+\frac{23}{16}-\left(5+\frac{27}{16}-\frac{19}{15}\right)+\left(\frac{-1}{2}\right)^2\)

\(=\frac{11}{15}+\frac{23}{16}-5-\frac{27}{16}+\frac{19}{15}+\frac{1}{4}\)

\(=\left(\frac{11}{15}+\frac{19}{15}\right)+\left(\frac{23}{16}-\frac{27}{16}\right)-5+\frac{1}{4}\)

\(=\frac{30}{15}-\frac{4}{16}-5+\frac{1}{4}\)

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

\(=-3\)

học tốt ngôlãmtân

30 tháng 11 2017

\(\frac{11}{15}+\frac{23}{16}-\left(5+\frac{27}{16}-\frac{19}{15}\right)+\left(-\frac{1}{2}\right)^2\) = \(\frac{11}{15}+\frac{23}{16}-5-\frac{27}{16}+\frac{19}{15}+\frac{1}{4}\)

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

Kb với mình nha!

11 tháng 12 2019

a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)

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

                                                     \(=0\)

b) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}=\left(\frac{11}{24}+\frac{13}{24}\right)+\left(-\frac{5}{41}-\frac{36}{41}\right)+0,5\)

                                                                    \(=1+\left(-1\right)+0,5\)

                                                                    \(=0,5\)

_Học tốt nha_

11 tháng 12 2019

a, \(\frac{15}{12}\)\(\frac{5}{13}\)\(\frac{3}{12}\)-\(\frac{18}{13}\)

\(\frac{5}{4}\)\(\frac{5}{13}\) - \(\frac{1}{4}\) - \(\frac{18}{13}\)

\(\left(\frac{5}{4}-\frac{1}{4}\right)\)\(\left(\frac{5}{13}-\frac{18}{13}\right)\)

= 1 - 1 = 0

b, \(\frac{11}{24}\)\(\frac{5}{41}\)\(\frac{13}{24}\)+ 0,5 - \(\frac{36}{41}\)

\(\left(\frac{11}{24}+\frac{13}{24}\right)\)\(\left(\frac{5}{41}+\frac{36}{41}\right)\)+ 0,5

= 1 - 1 + 0,5 = 0,5

c,  \(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)

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

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

\(\frac{11}{5}.\left[\left(-\frac{3}{4}-\frac{1}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)\right]\)

=  \(\frac{11}{5}.\left[\left(-1\right)+1\right]\)

= 0

d, \(\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)

\(9.\left(0,75-0,25\right)-2\)

= 9. 0,5 - 2 = 2,5

e, \(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)

\(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)

= -1 + 1 - \(\frac{1}{2}\)

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