47+145+124+100

=416 : 49

=416/49

giúp em với ạ

  49.(3022 + 3021) - 3022.49 + (100 - 49).3021

= 49.3022 + 49.3021 - 3022.49 + 100.3021 - 49.3021

= (49.3022 - 49.3022) + (49.3021 - 49.3021) + 100.3021

= 0 + 0 + 100.3021

= 302100

Theo đầu bài ta có:
\(49^5-49\)
\(\Leftrightarrow\left[\left(49^2\right)^2\cdot49\right]-49\)
\(\Leftrightarrow\left[\left(...01\right)^2\cdot49\right]-49\)
\(\Leftrightarrow\left[\left(...01\right)\cdot49\right]-49\)
\(\Leftrightarrow\left(...49\right)-49\)
\(\Leftrightarrow\left(...00\right)\)
Do 49⁵ - 49 có 2 chữ số 0 ở cuối nên chia hết cho 100. Vậy 49⁵ - 49 chia hết cho 100        ( đpcm )

=49[49^2+1].100.24

\(A=\left(\dfrac{1}{49}-\dfrac{1}{2^2}\right)\left(\dfrac{1}{49}-\dfrac{1}{3^2}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{100^2}\right)\)

\(=\left(\dfrac{1}{49}-\dfrac{1}{7^2}\right)\left(\dfrac{1}{49}-\dfrac{1}{2^2}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{100^2}\right)\)

\(=\left(\dfrac{1}{49}-\dfrac{1}{49}\right)\left(\dfrac{1}{49}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{10000}\right)\)

=0

125 = 100 + 25

21 = 22 - 1 

a) 100+25x100+1
b) 20+1x50-1

125.(100+1)=125.100+125.1=12500+125=12625

21.(50-1)=21.50-21.1=1050-21=1029

a, (367-24)-367=343-367

                      =(-24)

b,(17-600)+23=-583+23

                    =(-560)

c,-2016-(12-2016)=-2016-(-2004)

                        =(-2016)+2004

                       =(-12)

d, (18-29)-(24+18-29)=(-11)-13

                             =(-24)

e, -(13+49)+(13-100+49)=(-62)+(-38)

                                   =-100

f, -23+(100-57)=(-23)+43

                    =20

a) 125 . 101 = 125 . (100 + 1) = 125 . 100 + 125 . 1

= 12 500 + 125 = 12 625

b) 21 . 49 = 21 . (50 - 1) = 21 . 50 - 21 . 1

= 1 050 - 21 = 1 029