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25 tháng 7 2016

a ) \(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-12}{19}\)

\(=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{-12}{19}\)

\(=\frac{7}{19}.\frac{11}{11}+\frac{-12}{19}\)

\(=\frac{7}{19}.1+\frac{-12}{19}\)

\(=\frac{7}{19}+\frac{-12}{19}\)

\(=\frac{7+\left(-12\right)}{19}\)

\(=-\frac{5}{19}\)

b ) \(\frac{-7}{25}.\frac{39}{-14}.\frac{50}{78}=\frac{-7.39.50}{25.-14.78}=\frac{-1.1.2}{1.-2.2}=\frac{-2}{-4}=\frac{1}{2}\)

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

\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)

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

\(=\frac{3}{4}\)

25 tháng 7 2016

a)\(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-12}{19}=\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{7}{19}.\frac{-12}{7}=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}+-\frac{12}{7}\right)=\frac{7}{19}.\left(\frac{-5}{7}\right)=-\frac{5}{19}\)

b)\(\frac{-7}{25}.\frac{39}{-14}.\frac{50}{78}=\frac{\left(-7\right).39.50}{25.\left(-14\right).78}=\frac{\left(-7\right).3.13.2.5.5}{5.5.\left(-7\right).2.2.13.3}=\frac{1}{2}\)

c)\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}=\frac{2}{7}+\frac{1}{2}=\frac{11}{14}\)

25 tháng 7 2016

m) (\(\frac{-5}{12}\)+\(\frac{6}{11}\))+(\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\))

\(\frac{-5}{12}\)+\(\frac{6}{11}\)+\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\)

= (\(\frac{-5}{12}\)+\(\frac{5}{12}\))+(\(\frac{6}{11}\)+\(\frac{5}{11}\))+\(\frac{7}{17}\)

= 0+1+\(\frac{7}{17}\)

\(\frac{24}{17}\)

n) (\(\frac{9}{16}\)+\(\frac{8}{-27}\))+(1+\(\frac{7}{16}\)+\(\frac{-19}{27}\))

\(\frac{9}{16}\)+\(\frac{8}{-27}\)+1+\(\frac{7}{16}\)+\(\frac{-19}{27}\)

= (\(\frac{9}{16}\)+\(\frac{7}{16}\))+(\(\frac{8}{-27}\)+\(\frac{-19}{27}\))+1

= 1+(-1)+1

= 0+1

= 1

o) (6-2\(\frac{4}{5}\)).3\(\frac{1}{8}\)-1\(\frac{3}{5}\):\(\frac{1}{4}\)

= (6-\(\frac{14}{5}\)).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)

\(\frac{16}{5}\).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)

= 10-\(\frac{8}{5}\):\(\frac{1}{4}\)

= 10-\(\frac{32}{5}\)

\(\frac{18}{5}\)

CHÚC BẠN HỌC TỐT

 

9 tháng 4 2018

a) 2/7+-3/8+11/7+1/3+1/7+5/-8

=(2/7+11/7+1/7)+(3/8+-5/8)+1/3

=2+2+1/3

=4+1/3

=13/3

b) -3/8+12/25+5/-8+2/-5+13/25

=(-3/8+-5/8)+(12/25+13/25)+-2/5

=-1+1+-2/5

=0+-2/5

=-2/5

c)7/8+1/8*3/8+1/8*5/8

=7/8+1/8*(3/8+5/8)

=7/8+1/8*1

=7/8+1/8

=1

9 tháng 4 2018

a) 2/7+-3/8+11/7+1/3+1/7+5/-8

=(2/7+11/7+1/7)+(3/8+-5/8)+1/3

=2+2+1/3

=4+1/3

=13/3

b) -3/8+12/25+5/-8+2/-5+13/25

=(-3/8+-5/8)+(12/25+13/25)+-2/5

=-1+1+-2/5

=0+-2/5

=-2/5

c)7/8+1/8*3/8+1/8*5/8

=7/8+1/8*(3/8+5/8)

=7/8+1/8*1

=7/8+1/8

=1

9 tháng 2 2019

â, -4/9(7/15+8/15)=-4/9

b,-5/4(16/25+9/25)=-5/4

,..... 

dài quá mik làm ko hết 

hok tốt

6 tháng 5 2019

\(=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)

\(=\frac{7}{19}.1+\frac{12}{19}\)

\(=1\)

\(\frac{7}{19}\).\(\frac{8}{11}\)+\(\frac{7}{19}\).\(\frac{3}{11}\)+\(\frac{12}{19}\)

=\(\frac{7}{19}\).1+\(\frac{12}{19}\)

=1

hok tốt

18 tháng 8 2020

[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)

18 tháng 8 2020

\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)

9 tháng 8 2019

\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)

\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)

\(< =>\frac{-15+12-24}{63}\)

\(< =>\frac{-3}{7}\)

\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{7}\)

\(< =>\frac{29}{35}\)

\(bai2:\)

\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)

\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)

\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)

\(< =>x=\frac{3}{5}:\frac{-3}{4}\)

\(< =>x=\frac{-4}{5}\)

\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)

\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)

9 tháng 8 2019

Bài 1:

a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)                                 b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

 \(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\)                                       \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\) 

  \(=\frac{3}{7}.\frac{-9}{9}\)                                                                  \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(=\frac{-3}{7}\)                                                                           \(=\frac{7}{5}-\frac{4}{7}\)

                                                                                               \(=\frac{29}{35}\)

Bài 2:

a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)                                               b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

  \(\frac{-3}{4}x\)           \(=\frac{1}{5}+\frac{4}{10}\)                                     \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(\frac{-3}{4}x\)             \(=\frac{3}{5}\)                                            \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)     

         \(x\)              \(=\frac{3}{5}:\frac{-3}{4}\)                                        \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)                                         

         \(x\)              \(=\frac{4}{-5}\)                                                   \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)

                                                                                                             \(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\) 

                                                                                                                          \(x.\frac{10}{3}=\frac{13}{12}\) 

                                                                                                                                    \(x=\frac{13}{12}:\frac{10}{3}\) 

                                                                                                                                     \(x=\frac{13}{40}\)