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

\(\dfrac{3}{5}\times\dfrac{2}{7}-\dfrac{3}{5}\times\dfrac{4}{7} +\dfrac{3}{5}\)

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

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

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

\(=\dfrac{3}{5}\times\dfrac{5}{7}\)

\(=\dfrac{3}{7}\)

= 3/5 x 2/7 - 3/5 x 4/7 + 3/5 x 1

= 3/5 x ( 2/7 + 4/7 + 1 )

= 3/5 x 1

= 3/5

nhớ tích nha

3 \(\times\) \(\dfrac{4}{15}\) + 2 \(\times\) \(\dfrac{4}{15}\) - 5 \(\times\) \(\dfrac{4}{15}\)

= \(\dfrac{4}{15}\) \(\times\) ( 3 + 2 - 5)

= \(\dfrac{4}{15}\) \(\times\) 0

= 0

0 nha

[3+2-5]nhan 4/15

=0        nhan4/15

=0

=7/8.8/3.9/5.5

=7/3.9

=21

`7/8xx9/5xx8/3xx5=[7xx3xx3xx8xx5]/[8xx5xx3]=7xx3=21`

\(=\dfrac{7\times3\times3\times8\times5}{8\times5\times3}=\dfrac{21}{1}=21\)

(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

1, \(\dfrac{16\times25-22\times16}{7\times3+5\times7}=\dfrac{16\times\left(25-22\right)}{7\times\left(5+3\right)}=\dfrac{16\times3}{7\times8}\)

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

2,\(\dfrac{2001\times2003+2003\times2005}{2003\times4006}=\dfrac{2003\times\left(2001+2005\right)}{2003\times4006}=\dfrac{2003\times4006}{2003\times4006}=1\)

b ) Đặt \(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{101.103}\)

\(\Rightarrow A=\frac{5}{2}\left(\frac{5}{1}-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}+....+\frac{5}{101}-\frac{5}{103}\right)\)

\(\Rightarrow A=\frac{5}{2}\left(5-\frac{5}{103}\right)\)

Lời giải:
$2\times A=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{19\times 21}$
$2\times A=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+...+\frac{21-19}{19\times 21}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{21}$

$=1-\frac{1}{21}=\frac{20}{21}$

$\Rightarrow A=\frac{20}{21}: 2= \frac{10}{21}$