4/3+3-2/10=
* / là phần vd 4 phần ba
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\(\left(45\frac{1}{2}-2\frac{3}{8}\right)-\left(5\frac{5}{6}+6\frac{3}{4}\right)+\left(10\frac{2}{3}-5\frac{5}{8}\right)\)
\(=\left(\frac{91}{2}-\frac{19}{8}\right)-\left(\frac{35}{6}+\frac{27}{4}\right)+\left(\frac{32}{3}-\frac{45}{8}\right)\)
\(=\frac{345}{8}-\frac{151}{12}+\frac{121}{24}\)
\(=\frac{1035}{24}-\frac{302}{24}+\frac{121}{24}\)
\(=\frac{854}{24}\)\(=\frac{427}{12}\)
\(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)) +(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)):\(\frac{7}{5}\)
= \(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\))+(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)).\(\frac{5}{7}\)
=2.\(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\))
=\(\frac{10}{7}\).(\(\frac{21}{42}\)-\(\frac{8}{42}\)+\(\frac{24}{42}\))
=\(\frac{10}{7}\).\(\frac{37}{42}\)=\(\frac{185}{147}\)
2/3+1/3.(-4/9+5/6):7/12
=2/3+1/3.(-8/18+15/18):7/12
=2/3+1/3.7/18:7/12
=2/3+7/54:7/12
=2/3+7/54.12/7
=2/3+2/9
=6/9+2/9
=8/9
a, \(\dfrac{5}{6}\) = \(\dfrac{10}{12}\) = \(\dfrac{30}{36}\) b, \(\dfrac{40}{16}\) = \(\dfrac{10}{4}\) = \(\dfrac{200}{80}\)
c, \(\dfrac{9}{12}\) = \(\dfrac{3}{4}\) = \(\dfrac{54}{72}\) d, \(\dfrac{84}{21}\) = \(\dfrac{28}{7}\) = \(\dfrac{4}{1}\)
a)26,72+13/100 +21/40 = 5344/200+26/200+105/200=(5344+26+105)/200=219/8=27,375
c)3/4+15/40+69/300=150/200+75/200+46/200=(150+75+46)/200=271/200=1,355
a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)
\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)
hay \(x=-\dfrac{41}{12}\)
Vậy: \(x=-\dfrac{41}{12}\)
\(\frac{4}{3}+3-\frac{2}{10}=\frac{62}{15}\)