`25 xx 15 + 47 xx 95 + 25 xx 38 - 47 xx 70`

`= [ 25 xx ( 15 + 38)] + [ 47 xx ( 95 - 70)]`

`= [25 xx 53]+ [ 47 xx 25]`

`=25 xx 53 + 47xx 25`

`=25 xx ( 53 + 47)`

`=25 xx 100`

`=2500`

`#LeMichael`

`25.15+47.95+25.38-47.70`

`=25.(15+38)+47.(95-70)`

`=25.53+47.25`

`=25.(53+47)`

`=25.100`

`=2500`

27.(62+37)+73.(43+56)

= 27.99 + 73 . 99

= 99 .( 27+73)

= 99. 100

= 9900

`27.(62 + 37) + 73.(43+56)`

`=27.99 + 73 . 99`

`=99 . (27 + 73)`

`=99 . 100`

`=9900`

`#LeMichael`

\(B=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{99\times101}\)

\(B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{99}-\dfrac{1}{101}\)

\(B=1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(B=\dfrac{3-1}{1x3}+\dfrac{5-3}{3x5}+\dfrac{7-5}{5x7}+\dfrac{9-7}{7x9}+...+\dfrac{101-99}{99x101}\\ =\dfrac{3}{1x3}-\dfrac{1}{1x3}+\dfrac{5}{3x5}-\dfrac{3}{3x5}+\dfrac{7}{5x7}-\dfrac{5}{5x7}+\dfrac{9}{7x9}-\dfrac{7}{7x9}+...+\dfrac{101}{99x101}-\dfrac{99}{99x101}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ 1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\dfrac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.3^{14.2}}=\dfrac{11.3^{29}-3^{30}}{4.3^{28}}\)

\(=\dfrac{3^{28}\left(11.3-3^2\right)}{4.3^{28}}=\dfrac{33-9}{4}=\dfrac{24}{4}=6\)

Đề là gì bạn nhỉ?

cho mình đề bài ạ

Vì nếu tăng số lớn 124 đơn vị thì hiệu của 2 số đó sẽ tăng thêm 124 đơn vị

Vậy hiệu của 2 số đó khi số lớn không tăng thêm 124 đơn vị là:

150-124=26

Số lớn là : (2020+26):2=1023

Số bé là: 1023-26=997

Vậy số lớn là 1023, số bé là 997

\(B=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....0.....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =0\)

\(B=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{125}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....0.....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)=0\)

3/7,4/7,6/7
3/27,1/3,8/9
215/253,152/235,10/10,26/15,26/11