Cho biểu thức : A=999999999/2+9999999999/3+9999999999/6
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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
\(A=\frac{\text{9999999999}}{2}-\frac{\text{9999999999}}{3}-\frac{\text{9999999999}}{6}\)
\(A=\text{9999999999}\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(A=\text{9999999999}.0\)
\(A=0\)
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
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
\(A=\frac{999999999}{2}-\frac{999999999}{3}-\frac{999999999}{6}\)
\(=\frac{999999999\cdot3-999999999\cdot2-999999999}{6}\)
\(=\frac{999999999\cdot3-999999999\cdot\left(2+1\right)}{6}\)
\(=\frac{999999999\cdot3-999999999\cdot3}{6}=\frac{0}{6}=0\)
vậy A=0
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 ??????
\(\frac{9999999999}{2}+\frac{9999999999}{3}+\frac{9999999999}{6}\)
\(=9999999999\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)\)
\(=9999999999\cdot1\)
\(=9999999999\)