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\(C=\dfrac{1}{8}\times\dfrac{3}{2}-\dfrac{1}{8}\times\dfrac{1}{4}+\dfrac{3}{8}\times\dfrac{5}{8}\)
\(C=\dfrac{1}{8}\times\left(\dfrac{3}{2}-\dfrac{1}{4}\right)+\dfrac{3}{8}\times\dfrac{5}{4}\)
\(C=\dfrac{1}{8}\times\dfrac{5}{4}+\dfrac{3}{8}\times\dfrac{5}{4}\)
\(C=\left(\dfrac{1}{8}+\dfrac{3}{8}\right)\times\dfrac{5}{4}\)
\(C=\dfrac{1}{2}\times\dfrac{5}{4}\)
\(C=\dfrac{5}{8}\)
\(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
\(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{49}{30}\)
\(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{1}{5}\)
\(\dfrac{8}{5}x\dfrac{5}{8}=\dfrac{1}{1}=1\)
\(\dfrac{6}{7}x\dfrac{4}{7}=\dfrac{24}{49}\)
\(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}x\dfrac{5}{4}=\dfrac{1}{1}=1\)
\(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}x\dfrac{5}{5}=\dfrac{1}{1}=1\)
1) \(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{1+2}{3}=\dfrac{3}{3}=1\)
2) \(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{24+25}{30}=\dfrac{49}{30}\)
3) \(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{4-3}{5}=\dfrac{1}{5}\)
4) \(\dfrac{9}{8}-\dfrac{4}{2}=\dfrac{9}{8}-2=\dfrac{9}{8}-\dfrac{16}{8}=-\dfrac{7}{8}\)
5) \(\dfrac{8}{5}\times\dfrac{5}{8}=\dfrac{8\times5}{5\times8}=\dfrac{40}{40}=1\)
6) \(\dfrac{6}{7}\times\dfrac{4}{7}=\dfrac{6\times4}{7}=\dfrac{24}{7}\)
7) \(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{5}{4}=\dfrac{4\times5}{5\times4}=\dfrac{20}{20}=1\)
8) \(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}\times\dfrac{5}{5}=\dfrac{5\times5}{5\times5}=\dfrac{25}{25}=1\)
`a,3/10=0,3`
`3/100=0,03`
`4 25/100=4 1/4=4,25`
`2002/1000=2,002`
`b,1/4=0,25`
`3/5=0,6`
`7/8=0,875`
`1 1/2=1,5`
\(=\left(2+4+6+...+98\right)\left(6-6\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)\)
=0
a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)
\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)
\(3-a=\dfrac{3}{2}\)
\(a=3-\dfrac{3}{2}\)
\(a=\dfrac{6}{2}-\dfrac{3}{2}\)
\(a=\dfrac{3}{2}\)
b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)
\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)
\(0-a=\dfrac{7}{8}\)
\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)
c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)
\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)
\(\dfrac{8}{6}-a=\dfrac{1}{6}\)
\(a=\dfrac{8}{6}-\dfrac{1}{6}\)
\(a=\dfrac{7}{6}\)
a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)
a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)
a = 2 + 1 - 1 - \(\dfrac{1}{2}\)
a = 2 - \(\dfrac{1}{2}\)
a = \(\dfrac{3}{2}\)
b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
(3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)
a = - \(\dfrac{7}{8}\)
c, 2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a = \(\dfrac{1}{6}\)
a = 2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)
a = (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)
a = 1 + \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)
a = \(\dfrac{7}{6}\)
a) \(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\)
=> \(\dfrac{11}{4}\) \(-a+\dfrac{1}{4}=\dfrac{3}{2}\)
=> \(\dfrac{11}{4}-a\) = \(\dfrac{3}{2}-\dfrac{1}{4}\)
=> a = \(\dfrac{11}{4}-\dfrac{5}{4}\) =\(\dfrac{3}{2}\)
Vậy a = \(\dfrac{3}{2}\)
b) \(3\dfrac{1}{4}-a-1\dfrac{3}{4}=\dfrac{7}{8}\)
=> \(\dfrac{13}{4}-a-\dfrac{7}{4}=\dfrac{7}{8}\)
=> \(\dfrac{13}{4}-a=\dfrac{21}{8}\)
=> \(a=\dfrac{13}{4}-\dfrac{21}{8}=\dfrac{5}{8}\)
Vậy a = \(\dfrac{5}{8}\)
\(3\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{10}{3}+\dfrac{5}{6}=\dfrac{20}{6}+\dfrac{5}{6}=\dfrac{25}{6}\)
\(4\dfrac{5}{7}-2\dfrac{11}{21}=\dfrac{33}{7}-\dfrac{53}{21}=\dfrac{99}{21}-\dfrac{53}{21}=\dfrac{46}{21}\)
\(3\dfrac{1}{4}x\dfrac{2}{5}=\dfrac{13}{3}x\dfrac{2}{5}=\dfrac{26}{15}\)
\(4\dfrac{7}{8}:\dfrac{1}{4}=\dfrac{39}{8}.4=\dfrac{39}{2}\)
C
1/4 = 2/8