\(B=1000^9=999.1000^8+1000^8.\)

mà 999.1000⁸ > 999.999⁸ = 999⁹ ; 1000⁸ > 999⁸ 

\(\Rightarrow B>A\)

1 - \(\frac{999}{556}\) = \(\frac{-443}{556}\)

1 - \(\frac{1000}{557}\) = \(\frac{-443}{557}\)

Vì \(\frac{-443}{556}\) < \(\frac{-443}{557}\) nên \(\frac{999}{556}\) > \(\frac{1000}{557}\)

First we have :

\(\frac{999}{556}=\frac{999\cdot557}{556\cdot557}\)

Then : \(999\cdot557=999\cdot556+999\)

Next we have : \(1000\cdot556=999\cdot556+556\)

As you see : \(999\cdot556+556< 999\cdot556+999\)

So :\(\frac{999}{556}< \frac{1000}{557}\)

1/100 hả e hay là 1/10

Dạ 1/100

C= (1 - \(\frac{1}{2^2}\))+(1 - \(\frac{1}{3^2}\) )+(1 - \(\frac{1}{4^2}\))+.......+(1 - \(\frac{1}{100^2}\))

  =98 - (\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+........+\(\frac{1}{100^2}\))

=> C< 98            bn xem lai nha hinh nhu de sai  phai cong den \(\frac{9999}{10000}\)

Uk hinh nhu sai

Câu a:

CM (7^6 + 7^5 - 7^4) ⋮ 77

7^6 + 7^5 - 7^4

= 7^4.(7^2 + 7 - 1)

= 7^4.(49 + 7 - 1)

= 7^4.(56 - 1)

= 7^4.55

= 7^3.(7.11).5

= 7^3.77.5 ⋮ 77(đpcm)

Câu b:

Cm : (36^36 - 9^10) ⋮ 45

A = 36^36 - 9^10

A = 9^36.4^36 - 9^10

A = 9^10.(9^26.4^36 - 1)

A = 9^10.[(9^2)^13.(4^2)^18 - 1]

A = 9^10.[\(\overline{..1}^{13}\).\(\overline{..6}^{18}\) -1]

A = 9^10.[\(\overline{..6}-\overline{..1}\)]

A = 9^10.\(\overline{..5}\)

A ⋮ 9; 5

9 = 3^2; 5 = 5

BCNN(9; 5) = 45

A ∈ B(45) hay A ⋮ 45 (đpcm)