\(a,\dfrac{4}{5}+\dfrac{1}{4}\times\dfrac{2}{5}\)

\(=\dfrac{4}{5}+\dfrac{1}{10}\)

\(=\dfrac{8}{10}+\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

\(---\)

\(b,\dfrac{7}{12}+\dfrac{3}{44}:\dfrac{3}{8}-\dfrac{1}{4}\)

\(=\left(\dfrac{7}{12}-\dfrac{1}{4}\right)+\dfrac{3}{44}\cdot\dfrac{8}{3}\)

\(=\left(\dfrac{7}{12}-\dfrac{3}{12}\right)+\dfrac{8}{44}\)

\(=\dfrac{4}{12}+\dfrac{8}{44}\)

\(=\dfrac{44}{132}+\dfrac{24}{132}\)

\(=\dfrac{68}{132}\)

\(=\dfrac{17}{33}\)

#\(Toru\)

1 x 1 = 1

1 x 2 = 2

1 x 3 = 3

1 x 4 = 4

1 x 5 = 5

k mik nha bn

=5 kb nhe

* = dấu nhân

1 nhân 5 bằng 5

1

22/5

 a) X : 2 = 2/3 + 3/4 

     X : 2 = 17/12

     X      = 17/12 x 2 

     X       = 17/6 

b) X x 1/5 = 1 - 3/25

    X x 1/5 = 22/25

    X          = 22/25 : 1/5 

    X          = 22/5 

a) x:2= 17/12

x = 17/12 x 2

x= 17/6

b) X x 1/5 = 22/25

x = 22/15:1/5

x = 22/5

\(2^4\cdot5-\left[131-(13-4)^2\right]\)

\(=16\cdot5-\left[131-9^2\right]\)

\(=16\cdot5-\left[131-81\right]\)

\(=80-50=30\)

\(2^4.5-[131-(13-4)^2] \)

\(=16.5-[131-9^2]\)

\(=80-[131-81]\)

\(=80-50 \)

\(=30\)

2⁴.5[131-(13-4)² ]

=16.5-[131-9² ]

=80-[131-81]

=80-50

=30

\(a.\frac{1}{7}\times\frac{-3}{8}+\frac{-13}{8}==\frac{-3}{56}+\frac{-13}{8}=\frac{-3}{56}+\frac{-91}{56}=\frac{-94}{56}=\frac{-47}{28}\)

\(b.\frac{3}{5}\times\frac{13}{40}-\frac{1}{10}\times\frac{16}{23}=\frac{39}{200}-\frac{8}{115}=\frac{577}{4600}\)

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

\(=\left(\frac{-3}{4}+\frac{2}{5}\right)\times\frac{7}{3}+\left(\frac{3}{5}+\frac{1}{4}\right)\times\frac{7}{3}\)

\(=\frac{7}{3}\times\left(\frac{-3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{1}{4}\right)\)

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

\(=\frac{7}{3}\times\left(\frac{-2}{4}+1\right)\)

\(=\frac{7}{3}\times\frac{1}{2}\)

\(=\frac{7}{6}\)

\(d.\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{6}-\frac{5}{12}\right)\)

\(=\frac{7}{8}:\frac{7}{72}+\frac{7}{8}:\left(\frac{-1}{4}\right)\)

\(=\frac{7}{8}\times\frac{72}{7}+\frac{7}{8}\times-4\)

\(=\frac{7}{8}\times\left(\frac{72}{7}+\left(-4\right)\right)\)

\(=\frac{7}{8}\times\frac{44}{7}\)

\(=\frac{11}{2}\)

Lời giải:
a. 

$\frac{7}{4}+\frac{5}{-6}+\frac{21}{8}=\frac{42}{24}+\frac{-20}{24}+\frac{63}{24}=\frac{85}{24}$

b.

$\frac{4}{9}+\frac{-7}{12}+\frac{8}{15}$

$=\frac{80}{180}+\frac{-105}{180}+\frac{96}{180}=\frac{71}{180}$

c.

$=\frac{-1}{3}+\frac{5}{6}+\frac{19}{12}+2$
$=\frac{-4}{12}+\frac{10}{12}+\frac{19}{12}+2=\frac{25}{12}+2=\frac{49}{12}$

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