a.

b. Có 2 bạn cân nặng 45 kilogam.

Olm chào em, em làm như này là cưa đúng rồi, em nhé. 

\(b,\)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

bằng 2 nha bạn hải nam

gải ra hộ tớ

\(B=\frac{23^{41}+1}{23^{42}+1}\)

Vì B < 1

\(\Rightarrow B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23(23^{40}+1)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

P/s: Hoq chắc

ta có 

\(B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23\left(23^{40}+1\right)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

\(\Rightarrow B< A\)

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.....+\frac{1}{80}\)

\(=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+......+\frac{1}{80}\right)\)

\(>\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+.....+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\right)\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}\)

\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)

a:

N=15

c: Cân nặng trung bình là:

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

Ta có :

\(A=42\times75=\left(41+1\right)\times75=41\times75+75=41\times\left(74+1\right)+75=41\times74+116\)

\(B=41\times75+115\)

\(\Rightarrow A>B\)

A > B vì B = 59 . 43 mà A = 59 . 43 - 16. Suy ra A > B.