(Dấu . là dấu nhân)
a/\(\dfrac{2}{5}\cdot\dfrac{4}{3}-\dfrac{2}{5}:3\)
\(=\dfrac{2}{5}\cdot\dfrac{4}{3}-\dfrac{2}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(=\dfrac{2}{5}\cdot1\)
\(=\dfrac{2}{5}\)
b/\(\dfrac{2010}{2018}:\dfrac{1}{2}+\dfrac{7}{2018}:\dfrac{1}{2}\)
\(=\left(\dfrac{2010}{2018}+\dfrac{7}{2018}\right):\dfrac{1}{2}\)
\(=\dfrac{2017}{2018}:\dfrac{1}{2}\)
\(=\dfrac{2017}{2018}\cdot2\)
\(=\dfrac{2017}{1009}\)

a, \(\dfrac{2}{5}\) \(\times\) \(\dfrac{4}{3}\) - \(\dfrac{2}{5}\) : 3

=  \(\dfrac{2}{5}\) \(\times\) \(\dfrac{4}{3}\) - \(\dfrac{2}{5}\) \(\times\) \(\dfrac{1}{3}\)

= \(\dfrac{2}{5}\) \(\times\) ( \(\dfrac{4}{3}\) - \(\dfrac{1}{3}\))

= \(\dfrac{2}{5}\)  \(\times\) 1

= \(\dfrac{2}{5}\) 

b, \(\dfrac{2010}{2018}\) : \(\dfrac{1}{2}\) + \(\dfrac{7}{2018}\) : \(\dfrac{1}{2}\) + \(\dfrac{1}{2018}\) : \(\dfrac{1}{2}\)

= \(\dfrac{2010}{2018}\) \(\times\) \(\dfrac{2}{1}\) + \(\dfrac{7}{2018}\) \(\times\) \(\dfrac{2}{1}\) + \(\dfrac{1}{2018}\) \(\times\) \(\dfrac{2}{1}\)

= \(\dfrac{2}{1}\) \(\times\) ( \(\dfrac{2010}{2018}\) + \(\dfrac{7}{2018}\) + \(\dfrac{1}{2018}\))

= 2 \(\times\) \(\dfrac{2018}{2018}\)

= 2 \(\times\) 1

= 2

a, 13 \(\times\) 15 - 150 + 97 \(\times\)15

13 \(\times\) 15 - 15 \(\times\) 10 + 97 \(\times\) 15

= 15 \(\times\) ( 13 - 10 + 97)

= 15 \(\times\) ( 3 + 97)

= 15 \(\times\) 100

=1500   

b, \(\dfrac{2016}{2015}\) \(\times\) \(\dfrac{4}{7}\) - \(\dfrac{4}{7}\) \(\times\) \(\dfrac{1}{2015}\) + \(\dfrac{3}{7}\)

= \(\dfrac{4}{7}\) \(\times\) ( \(\dfrac{2016}{2015}\) - \(\dfrac{1}{2015}\)) + \(\dfrac{3}{7}\) 

= \(\dfrac{4}{7}\) \(\times\) \(\dfrac{2015}{2015}\) + \(\dfrac{3}{7}\)

= \(\dfrac{4}{7}\) + \(\dfrac{3}{7}\)

= \(\dfrac{7}{7}\)

= 1

a, \(13\) \(\times\) 15 - 150 + 97 \(\times\) 15

= 15 \(\times\) ( 13 - 10 + 97)

= 15 \(\times\) ( 3 + 97)

= 15 \(\times\) 100

= 1500  

b, \(\dfrac{2016}{2015}\) \(\times\) \(\dfrac{4}{7}\) - \(\dfrac{4}{7}\) \(\times\) \(\dfrac{1}{2015}\) + \(\dfrac{3}{7}\)

= \(\dfrac{4}{7}\) \(\times\) ( \(\dfrac{2016}{2015}\) - \(\dfrac{1}{2015}\)) + \(\dfrac{3}{7}\)

= \(\dfrac{4}{7}\) \(\times\) \(\dfrac{2015}{2015}\) + \(\dfrac{3}{7}\)

= \(\dfrac{4}{7}\) + \(\dfrac{3}{7}\)

= \(\dfrac{7}{7}\)

= 1

4/5x3/7+4/5+6/7x6/7-4/5x4/14

= ( 3/7 + 6/7 + 4/14) x 4/5 

= 11/7 x 4/5 

= 44/35

44/35

4/5x3/7+4/5+6/7x6/7-4/5x4/14 ⇒ ( 3/7 + 6/7 + 4/14) x 4/5 ⇒ 11/7 x 4/5 ⇒ 44/35

\(\frac{3}{4}x\frac{4}{5}:\frac{3}{5}x\frac{6}{7}:\frac{1}{7}x\frac{5}{6}\)\(=\)\(\frac{3}{4}x\frac{4}{5}x\frac{5}{3}x\frac{6}{7}x\frac{7}{1}x\frac{5}{6}\)\(=\)\(\frac{3x4x5x6x7x5}{4x5x3x7x1x6}\)\(=\)\(\frac{\left(3\right)x\left(4\right)x\left(5\right)x\left(6\right)x\left(7\right)x5}{\left(3\right)x\left(4\right)x\left(5\right)x\left(6\right)x\left(7\right)x1}\)\(=\)\(\frac{5}{1}\)\(=\)\(5\).

5

`a)17/7-3/8+4/7-3/5`

`=(17/7+4/7)-(3/8+3/5)`

`=21/7-(15/40+24/40)=3-39/40=120/40-39/40=81/40`

`b)15/11xx2001/2005-3/11xx2001/2005-1/11xx2001/2005`

`=2001/2005xx(15/11-3/11-1/11)`

`=2001/2005xx11/11`

`=2001/2005`

3/4 x 4/5 : 3/5 x 6/7 : 1/7 x 5/6

= 3/4 x 4/5 x 5/3 x 6/7 x7/1 x 5/6

= 5

\(A=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{99.101}\right)\)

\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{7}{2}.\frac{100}{101}\)

\(=\frac{350}{101}\)

\(A=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{99.101}\right)\)

\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{7}{2}.\frac{100}{101}\)

\(=\frac{350}{101}\)

\(A=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{99.101}\right)\)

\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{7}{2}.\frac{100}{101}\)

\(=\frac{350}{101}\)