(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

\(2018\times\frac{3}{4}+2018\times\frac{1}{4}-2018\)

\(=2018\times\left(\frac{3}{4}+\frac{1}{4}-1\right)\)

\(=2018\times0=0\)

B-a:tương tự bài A

b,\(\frac{14}{18}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)

\(=\frac{7}{9}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)

\(=\frac{7}{9}\times\left(\frac{8}{5}-\frac{3}{5}\right)\)

\(=\frac{7}{9}\times1=\frac{7}{9}\)\

thanks

B-a,\(\frac{3\times125+3\times125}{6\times43+6\times57}=\frac{2\times3\times125}{6\times\left(43+57\right)}\)

\(=\frac{3\times250}{6\times100}=\frac{5}{2\times2}=\frac{5}{4}\)

thanks

A) 2018*3/4+2018*1/4-2018

=2018*(3/4+1/4)-2018

=2018*4/4-2018

=2018*1-2018

=2018-2018

=0

bn thông cảm mình có việc bận,chút nữa mình trả lời câu B

2018x36+2018+2018x63

=2018x(36+1+63)

=2018x100

=201800

2018 x 36 + 2018 + 2018 x 63

= 2018 x 36 + 2018 x 1 + 2018 x 63

= 2018 x ( 36 + 1 + 63 )

=2018 x 100

= 201800

^HT^

2018 nhân 36 +2018 nhân 1 +2018 nhân 63

=2018 nhân [ 36 +63] +1

=2018 nhân 100

= 201800

\(\frac{2017}{2018}\)x\(\frac{7}{8}\)+\(\frac{2017}{2018}\)x\(\frac{3}{8}\)-\(\frac{2017}{2018}\)x\(\frac{1}{4}\)

= \(\frac{2017}{2018}\)x  (\(\frac{7}{8}\)+\(\frac{3}{8}\)-\(\frac{1}{4}\))

= \(\frac{2017}{2018}\)x  ( \(\frac{10}{8}\)- \(\frac{1}{4}\))

= \(\frac{2017}{2018}\)x ( \(\frac{10}{8}\)- \(\frac{2}{8}\))

= \(\frac{2017}{2018}\)x  1

= \(\frac{2017}{2018}\)

Chúc em học tốt nhé :>

=2017/2018*(7/8+3/8)-2017*1/4

=2017/2018*5/4+2017*-1/4

=2017/2018*(5/4-1/4)

=2017/2018*1

=2017/2018

Bài 1:a, Tính bằng cách thuận tiện nhất :

2018 x 32 + 8072 : 4 x 23 + 4036 : 2 x 25 + 2018 +2018 x 19

=2018 x 32 +  2018  x 23 + 2018 x 25 + 2018 x 1 + 2018 x 19

= 2018 x ( 32 + 23 + 25 + 1 + 19 )

= 2018 x 100 

=201800

Bài 2 :a, Tính bằng cách thuật tiện :

3 x 4/15 + 2 x 4/15 - 5 x 4/15

= ( 3 x 2 - 5 ) x 4/15

=     0      x 4/15

=      0

Bài 

\(= 2018 × 100 \)

\(= 201800 \)

b ) Số hoa còn lại là :

\(26:\dfrac{2}{5}=65bông\)

Tất cả có số bông hoa là :

\(65+5=71bông\)

bài 3 :

\(a\))\(3\text{×}\dfrac{4}{15}+2\text{×}\dfrac{4}{15}-5\text{×}\dfrac{4}{15}\)

\(=\dfrac{4}{15}\text{×}\left(3+2-5\right)\)

\(=\dfrac{4}{15}\text{×}0\)

\(=0\)

\(x = 42\)

Vậy