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2 tháng 7 2019

a, \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)

\(=\frac{-5}{9}.\left(\frac{3}{10}-\frac{4}{10}\right)\)

\(=\frac{-5}{9}.\frac{-1}{10}\)

\(=\frac{5}{90}\)

\(=\frac{1}{18}\)

b,\(\frac{2}{3}+\frac{-1}{3}+\frac{7}{15}\)

\(=\frac{10}{15}-\frac{5}{15}+\frac{7}{15}\)

\(=\frac{12}{15}\)

\(=\frac{4}{5}\)

c, \(\frac{3}{8}.3\frac{1}{3}\)

\(=\frac{3}{8}.\frac{10}{3}\)

\(=\frac{10}{8}\)

\(=\frac{5}{4}\)

d, \(\frac{-3}{5}+0,8.\left(-7\frac{1}{2}\right)\)

\(=\frac{-3}{5}+\frac{4}{5}.\frac{-15}{2}\)

\(=\frac{-3}{5}+\frac{-60}{10}\)

\(=\frac{-3}{5}+\frac{-30}{5}\)

\(=\frac{-33}{5}\)

e, \(\frac{2}{5}.8\frac{1}{3}+1\frac{2}{3}.\frac{2}{5}\)

\(=\frac{2}{5}.\left(8\frac{1}{3}+1\frac{2}{3}\right)\)

\(=\frac{2}{5}.10\)

\(=4\)

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

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

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

\(=-6\)

~Study well~

#KSJ

4 tháng 10 2021

yutyugubhujyikiu

1 tháng 2 2020

\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)

...
Đọc tiếp

\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)

\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)

\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)

\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)

\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)

\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)

\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)

\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)

\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)

\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)

\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)

\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)

\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)

\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)

\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)

\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)

\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)

\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)

TRÌNH BÀY GIÚP MÌNH NHA 

0
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023

a)

\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)

b)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\  = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)

Bài 1:

a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)

\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)

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

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

b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)

\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)

\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)

\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)

c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)

\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)

\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)

d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)

\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)

\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)

\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)

e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)

\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)

\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)

\(=\frac{11}{12}\)

f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)

\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)

\(=\frac{1}{5}\)

...
Đọc tiếp

\(3\frac{1}{2}-4\frac{2}{3}+\left[\frac{3}{4}-2\frac{1}{3}\right]-\left(\frac{5}{6}-\frac{7}{4}\right)+5\frac{1}{2}-3\)

\(2\frac{2}{3}-1\frac{2}{5}+1\frac{3}{10}-\left(\frac{2}{5}-\frac{5}{6}\right)+\frac{4}{15}-1\frac{1}{3}\)

\(\left[2\frac{1}{3}-1\frac{4}{3}\right]-\left(\frac{5}{4}-\frac{7}{12}+\frac{-11}{6}\right)+\frac{4}{3}-\frac{3}{4}\)

\(-3\frac{3}{2}+5\frac{4}{3}-\left(\frac{7}{6}-1\frac{3}{4}\right)+\left[\frac{2}{3}-2\frac{1}{4}\right]\)

\(2\frac{2}{3}-\frac{5}{12}-\left(1\frac{3}{4}-2\frac{1}{4}\right)-\left[1-1\frac{1}{6}\right]+\left[\frac{-5}{3}\right]\)

\(1\frac{1}{3}-5\frac{1}{2}-\left[\frac{5}{6}-2\frac{2}{3}\right]+\left[\frac{7}{12}-\frac{5}{6}\right]\)

\(\frac{8}{15}-\left(\frac{2}{5}-3\frac{1}{3}+\left[\frac{-5}{6}\right]\right)+\left[\frac{1}{2}-\frac{4}{5}\right]-\left(\frac{1}{6}-1\frac{1}{3}\right)\)

\(-2\frac{3}{2}+\left[\frac{5}{6}-1\frac{1}{3}\right]-\left(\frac{5}{12}-\frac{7}{6}\right)+\left[\frac{4}{3}-3\frac{1}{4}\right]\)

\(\frac{9}{10}-1\frac{2}{5}-\left(\frac{5}{6}-3\frac{1}{2}\right)-\left[2\frac{1}{4}-5\frac{2}{36}\right]-\left[1-2\frac{1}{15}\right]\)

\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)

\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)

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

\(2\frac{1}{3}-\left(5\frac{1}{2}-2\frac{2}{3}\right)+\left[1\frac{1}{6}-2\frac{1}{2}\right]-\frac{5}{12}+\left(\frac{1}{4}-\frac{1}{8}\right)\)

 

 

 

 

 

 

 

 

2
19 tháng 6 2018
  1. ​29/15
  2. 23
  3. 23/12
  4. 5/6
  5. 5/4
  6. -31/12
  7. 31/6
  8. -13/3
  9. 1087/180
  10. 1/6
  11. 1/6
  12. 2
  13. -67/24
11 tháng 4 2022
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