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\(\left(\dfrac{3}{15}-x\right)\cdot\dfrac{1}{3}=\dfrac{2}{5}\)

=>\(\dfrac{1}{5}-x=\dfrac{2}{5}:\dfrac{1}{3}=\dfrac{2}{5}\cdot3=\dfrac{6}{5}\)

=>\(x=\dfrac{1}{5}-\dfrac{6}{5}=-\dfrac{5}{5}=-1\)

1 tháng 2 2019

a) \(x-\frac{10}{3}=\frac{7}{15}\cdot\frac{3}{5}\)                          b) \(x+\frac{3}{22}=\frac{27}{121}\cdot\frac{11}{9}\)

\(\Leftrightarrow x-\frac{10}{3}=\frac{7}{25}\)                                \(\Leftrightarrow x+\frac{3}{22}=\frac{3}{11}\)

\(\Rightarrow x=\frac{7}{25}+\frac{10}{3}\)                                   \(\Rightarrow x=\frac{3}{11}-\frac{3}{22}\)

       \(x=\frac{271}{75}\)                                                  \(x=\frac{3}{22}\)

c) \(\frac{8}{23}.\frac{46}{24}-x=\frac{1}{3}\)                    d) \(1-x=\frac{49}{65}.\frac{5}{7}\)

        \(\Leftrightarrow\frac{2}{3}-x=\frac{1}{3}\)                      \(\Leftrightarrow1-x=\frac{7}{13}\)

                      \(\Rightarrow x=\frac{2}{3}-\frac{1}{3}\)                       \(\Rightarrow x=1-\frac{7}{13}\)

                           \(x=\frac{1}{3}\)                                         \(x=\frac{6}{13}\)

                                 

25 tháng 6 2017

\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)

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

\(x=1\div\left(-\frac{3}{5}\right)\)

\(x=-\frac{5}{3}\)

\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)

\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)

\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)

\(-\frac{4}{21}\cdot x=\frac{3}{5}\)

\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)

\(x=-\frac{63}{20}\)

25 tháng 6 2017

\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)

\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)

\(\frac{5}{7}\cdot x=1\)

\(x=1\div\frac{5}{7}\)

\(x=\frac{7}{5}\)

\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)

\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)

\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)

\(\frac{17}{33}\cdot x=\frac{9}{14}\)

\(x=\frac{9}{14}\div\frac{17}{33}\)

\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)

11 tháng 7 2019

1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)

= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)

= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)

= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))

= \(\frac{284}{8}\). \(\frac{-9}{284}\)

= \(\frac{-9}{8}\)

25 tháng 4 2018

a) x = 99/20

b) x = 7

c) x = 2

( chỉ lm đc đến đó thui nk )

24 tháng 7 2019

a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)

\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)

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

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

b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)

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

\(=2-1\)

\(=1\)

c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)

\(=\frac{-1}{4}+\frac{-16}{11}\)

\(=\frac{-75}{44}\)

d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)

\(=\frac{-6}{11}-\frac{3}{22}\)

\(=\frac{15}{22}\)

e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\) 

\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(=\frac{1}{3}-\frac{1}{15}\)

\(=\frac{4}{15}\)

11 tháng 4 2019

\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)

\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)

\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)

\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)

\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)

Đến đây là tính dễ rồi :v

\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)

\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)

\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)

\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)

\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)

\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)

Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)

\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)