\(2019+2019+2019+2019+2019\times7-2019\\ =2019\times\left(1+1+1+1+7-1\right)\\ =2019\times10\\ =20190\)

Tính bằng cách thuận tiện nhất.

2019 + 2019 +2019 + 2019 + 2019 x 7 - 2019

= 2019 x ( 7 +1 +1+1 - 1 ) = 2019 x 10 = 20190

= 20190

( - 2019 - 2019 - 2019 - 2019 ) x ( -25 ) 

= (-4 ) x 2019 x ( - 25 ) 

= ( -4 ) x ( -25 ) 2019 

= 100 x 2019 

= 201900 

học tốt 

201900

(1/2019)^2020 . 2019^2019

= \(\frac{1}{2019^{2020}}\cdot2019^{2019}\)

= \(\frac{2019^{2019}}{2019^{2020}}\)

=1/2019

vậy.......

hok tốt

Thanks

Trả lời :...............................................

\(\frac{4078379}{4078379}\)

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k nhé Kim Râu La

\(\frac{2019}{1\times2}+\frac{2019}{2\times3}+\frac{2019}{3\times4}+...+\frac{2019}{2018\times2019}\)

\(=2019\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2018\times2019}\right)\)

\(=2019\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2019\left(1-\frac{1}{2019}\right)\)

\(=2019\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)

\(=2019\times\frac{2018}{2019}\)\(=\frac{2019\times2018}{2019}=2018\)

a) (4+4)x(2019 +2019 +.....+2019)

= 8 x 125 x 2019

= (8 x 125) x 2019

= 1000 x 2019

= 2019000

b) ( 42+42+...+42)+(125+125+....+125)

= 42 x 125 + 58 x 125

= 125 x ( 42 + 58)

= 125 x 100

= 12500

 42+42+...+42)+(125+125+....+125)

= 42 x 125 + 58 x 125

= 125 x ( 42 + 58)

= 125 x 100

= 12500

\(2019\times58+2019\times42+78\)

\(=2019\times\left(58+42\right)+78\)

\(=2019\times100+78\)

\(=201900+78\)

\(=201978\)

2019 x  58 + 2019 x 42 + 78

= 2019 x ( 58 + 42 ) + 78

= 2019 x 100 + 78

= 201900 + 78 = 201978

A = 2019  x 58 + 2019 x 42 + 78

A = (2019 x 58 + 2019 x 42) + 78

A = 2019 x ( 58 + 42) + 78

A = 2019 x 100 + 78

A = 201900 + 78

A = 201978

\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+....+\frac{2019}{2018.2019}\)

   \(=\frac{2019}{1}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2018.2019}\right)\)

   \(=\frac{2019}{1}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)

   \(=\frac{2019}{1}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\right)\)

   \(=\frac{2019}{1}.\left(1-\frac{1}{2019}\right)\)

   \(=\frac{2019}{1}.\frac{2018}{2019}\)

   \(=2018\)

\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+\frac{2019}{20}+\frac{2019}{30}+\frac{2019}{2018.2019}\)

\(A=\frac{2019}{1.2}+\frac{2019}{2.3}+\frac{2019}{3.4}+\frac{2019}{4.5}+\frac{2019}{5.6}+...+\frac{2019}{2018.2019}\)

\(A=2019.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)

\(A=2019.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(A=2019.\left(1-\frac{1}{2019}\right)\)\(=2019.\frac{2018}{2019}=2018\)

Vậy A = 2018 

-Dấu " . " là dấu nhân.