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

\(\frac{9999999999}{2}+\frac{9999999999}{3}+\frac{9999999999}{6}\)

\(=9999999999\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)\)

\(=9999999999\cdot1\)

\(=9999999999\)

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

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

\(\Rightarrow A>0\left(đpcm\right)\)

19999999998

= 19999999998

19999999998

9999999999 + 9999999999 = 1999999998

Mk tính nhẩm đấy

9999999999+9999999999=19999999998

= 19999999998

Bằng 19 999 999 998

Bằng 1 999 999 998

=19999999998 nhé

chúc bạn học tốt

9999999999x9999999999=99999999980000000001

9999999999 x 9999999999 = 9999999999^2

9999999999 x 9999999999 =9999999999^2

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 ??????