\(b,\)Vì p là SNT > 3 => p có dạng : 3k + 1 ; 3k + 2 ( k thuộc N)

Với p = 3k + 1

\(=>\left(3k+2\right)\left(3k\right)⋮3\)(1)

Với p = 3k + 2

\(=>\left(3k+3\right)\left(3k+1\right)=3\left(k+1\right)\left(3k+1\right)⋮3\)(2)

Từ (1) và (2) => ĐPCM

\(A=\frac{11^{40}+1}{11^{43}+1}\)

\(11^3A=\frac{11^3\left(11^{40}+1\right)}{11^{43}+1}=\frac{11^{43}+1331}{11^{43}+1}=\frac{11^{43}+1+1330}{11^{43}+1}=\frac{11^{43}+1}{11^{43}+1}+\frac{1330}{11^{43}+1}=1+\frac{1330}{11^{43}+1}\)

\(B=\frac{11^{41}+1}{11^{44}+3}\)

\(11^3B=\frac{11^3\left(11^{41}+1\right)}{11^{44}+3}=\frac{11^{44}+1331}{11^{44}+3}=\frac{11^{44}+3+1328}{11^{44}+3}=\frac{11^{44}+3}{11^{44}+3}+\frac{1328}{11^{44}+3}=1+\frac{1328}{11^{44}+3}\)

Ta có: \(\frac{1330}{11^{43}+1}>\frac{1330}{11^{44}+3}>\frac{1328}{11^{44}+3}\)

\(\Rightarrow\frac{1330}{11^{43}+1}>\frac{1328}{11^{44}+3}\)

\(\Rightarrow1+\frac{1330}{11^{43}+1}>1+\frac{1328}{11^{44}+3}\)

\(\Rightarrow11^3A>11^3B\)

\(\Rightarrow A>B\)

Vậy \(A>B\)

P/s: Hoq chắc :<

\(4S=4+4^2+4^3+4^4+...+4^{100}\)

\(3S=4S-S=4^{100}-1\Rightarrow3S+1=4^{100}\)

Ta có \(32^{20}=\left(2^5\right)^{20}=2^{100}\)

\(\Rightarrow4^{100}>2^{100}\Rightarrow3S+1>32^{20}\)

Bài 1:

\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)

Bài 2:

\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)

Bài 1 :

\(2^{49}=\left(2^7\right)^7=128^7\)

\(5^{21}=\left(5^3\right)^7=125^7\)

mà \(125^7< 128^7\)

\(\Rightarrow2^{49}>5^{21}\)

Bài 2 :

a) \(S=1+3+3^2+3^3+...3^{99}\)

\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)

\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

b) \(S=1+4+4^2+4^3+...4^{62}\)

\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)

\(\Rightarrow S=21+4^3.21+...4^{60}.21\)

\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)

\(\Rightarrow dpcm\)

B = (4^1 + 4^2) + (4^3 +4^4) + ... + (4^299 + 4^300)

= 4(1+4)+4^3(1+4)+...+4^299(1+4)

= 4.5+4^3 .5 +...+4^299. 5

= 5.(4+4^3+...+4^299) chia hết cho 5

\(B=4^1+4^2+4^3+4^3+...+4^{300}\\=(4+4^2)+(4^3+4^4)+(4^5+4^6)+...+(4^{299}+4^{300})\\=4\cdot(1+4)+4^3\cdot(1+4)+4^5\cdot(1+4)+...+4^{299}\cdot(1+4)\\=4\cdot5+4^3\cdot5+4^5\cdot5+...+4^{299}\cdot5\\=5\cdot(4+4^3+4^5+...+4^{299})\)

Vì \(5\cdot(4+4^3+4^5+...+4^{299}) \vdots 5\)

nên \(B \vdots 5\)

a:

N=15

c: Cân nặng trung bình là:

\(\dfrac{39\cdot1+40\cdot4+41\cdot3+42\cdot4+43+45\cdot2}{15}\simeq41,5\left(kg\right)\)

a.

b. Có 2 bạn cân nặng 45 kilogam.

41 + 42 + 43 + 44 - 21 - 22 - 23 - 24

= (41 - 21) + (42 - 22) + (43 - 23) + (44 - 24)

= 20 + 20 + 20 + 20

= 20 x 4

= 80

tk nha. chúc bn học giỏi :)

= 80 nha

TK MK NHA CÁC BẠN MK ĐG BỊ ÂM ĐIỂM

Số số hạng của tổng:

60 - 41 + 1 = 20

Ta có:

1/41 + 1/42 + 1/43 + ... + 1/60 > 1/60 + 1/60 + 1/60 + ... + 1/60 (20 số 1/60)

= 20/60

= 1/3

Vậy 1/41 + 1/42 + 1/43 + ... + 1/60 > 1/3