Nhân xét biểu thức A, ta thấy:

   \(\frac{9999999999}{2}>\frac{9999999999}{3}>\frac{9999999999}{6}>0\)

=> \(A>0\left(đpcm\right)\)

ta có 9999999999/2=9999999999*3/2*3

9999999999/3=9999999999*2/3*2

suy ra 9999999999*3/2*3 - 9999999999*2/3*2=9999999999*3-9999999999*2/6=9999999999/6

suy ra A=9999999999/6-9999999999/6=0

vậy A=0

\(A=9999999999.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(A=9999999999.0=0\)

Vậy A=0

\(A=\frac{\text{9999999999}}{2}-\frac{\text{9999999999}}{3}-\frac{\text{9999999999}}{6}\)

\(A=\text{9999999999}\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(A=\text{9999999999}.0\)

\(A=0\)

Vậy A = B 

A = 0 nhé bạn tôi thề là đúng luôn

bạn ko cần cảm ơn đâu cho 1 k là ok

a)

\(A=\frac{x}{y}\Leftrightarrow n-2\ne0\Leftrightarrow n\ne2\)

b)

A là số nguyên khi \(n-2\inƯ_{-5}\)

\(\Rightarrow n-2\in\left\{1;5;-1;-5\right\}\)

\(\Rightarrow n\in\left\{3;8;1;-3\right\}\)

Vậy \(n\in\left\{3;8;1;-3\right\}\)

Đặt BT là B

\(\Rightarrow B=3\left(1+3^2+3^2+3^3\right)+.......+3^{97}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow B=3.40+....+3^{97}.40\) chia hết cho 40

=> B chia hết cho 40

\(A=\frac{999999999}{2}-\left(\frac{999999999.2}{9}-\frac{999999999}{6}\right)\)

\(A=\frac{999999999}{2}-\frac{999999999}{6}\)

Vì \(\frac{999999999}{2}>\frac{999999999}{6}\)

\(\Rightarrow\frac{999999999}{2}-\frac{999999999}{6}>0\)

\(\Rightarrow A>0\left(đpcm\right)\)

9 . 1 + 9. 11 + 9.111 + 9.1111 + 9.11111 + 9.111111 + 9.1111111 + 9.11111111 + 9.111111111 + 9.1111111111

= (10-1) + (100 -1) + (1000-1) + (10000-1) + ............+ (10000000000-1)

= ( 10+100+1000+10000+..........+10000000000) - (1+1+1+1+..........+1+1)

=11111111110 - 1.10

=1111111100

= (10-1) + (100 -1) + (1000-1) + (10000-1) + ............+ (10000000000-1)

= ( 10+100+1000+10000+..........+10000000000) - (1+1+1+1+..........+1+1)

=11111111110 - 1.10

=1111111100

11111111110-1.10

= 1111111100

\(A=\frac{999999999}{2}-\frac{999999999}{3}-\frac{999999999}{6}\)

   \(=\frac{999999999\cdot3-999999999\cdot2-999999999}{6}\)

   \(=\frac{999999999\cdot3-999999999\cdot\left(2+1\right)}{6}\)

   \(=\frac{999999999\cdot3-999999999\cdot3}{6}=\frac{0}{6}=0\)

vậy A=0

tôi thề là A = 0

tôi vừa tính xong

đúng hơn là tính máy tính

A=999999999.(\(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\))=999999999.0=0=A

9999999999 + 9999999999 = 1999999998

Mk tính nhẩm đấy

9999999999+9999999999=19999999998

= 19999999998

12/25

3623/50000

64

mik làm đầu tiên

k mik nhé

4 : 23  x 3

= 4/23  x 3

= 12/23

6 : 1 + ? = 70

6 + ? = 70

? = 70 - 6

? = 64

tk mk nha

mà vô '' khách sạn '' làm gì thế bạn ??????