Ta có: \(\frac{1}{31}<\frac{1}{30};\frac{1}{32}<\frac{1}{30};...;\frac{1}{40}<\frac{1}{30}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}<\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (1)

Ta có: \(\frac{1}{41}<\frac{1}{40};\frac{1}{42}<\frac{1}{40};\ldots;\frac{1}{50}<\frac{1}{40}\)

Do đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{50}<\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{10}{40}=\frac14\) (2)

Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{60}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{60}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{10}{50}=\frac15\) (3)

Từ (1),(2),(3) suy ra \(\left(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{60}\right)<\frac13+\frac14+\frac15\)

=>\(A<\frac{47}{60}<\frac{48}{60}=\frac45\)

\(B=3^0+3^1+3^2...+3^{100}\)

\(=3^0\times\left(1+3^1+3^2\right)+3^3\times\left(1+3^1+3^2\right)+...+3^{98}\times\left(1+3^1+3^2\right)\)

\(=3^0\times13+3^3\times13+...+3^{98}\times13\)

\(=13\times\left(3^0+3^3+...+3^{98}\right)⋮13\)

$B=3^0+3^1+3^2...+3^{100}$

$=3^0\times \left(1+3^1+3^2\right)+3^3\times \left(1+3^1+3^2\right)+...+3^{98}\times \left(1+3^1+3^2\right)$

$=3^0\times 13+3^3\times 13+...+3^{98}\times 13$

$=13$

ta có: \(\frac{31+32+35}{34}=\frac{31}{34}+\frac{32}{34}+\frac{35}{34}.\)

mà \(\frac{31}{32}>\frac{31}{34};\frac{32}{33}>\frac{32}{34}\)

\(\Rightarrow\frac{31}{32}+\frac{32}{33}+\frac{35}{34}>\frac{31}{34}+\frac{32}{34}+\frac{35}{34}=\frac{31+32+35}{34}\)

Mk giải 1 phần phần còn lại tương tự nhé ^^

1)   A = 25 . 33 - 10= ( 26 - 1 ). 33 -10= 26.33 -33 - 10= 26.33 -43

B= 31.26 + 10 = ( 33-2).26 +10= 26.33 - 2.26 + 10= 26.33  - 42

=> A>B

ko bt làm thì có :))

b) B\(=\)(3+3²+3³)+...+(3⁵⁸+3⁵⁹+3⁶⁰)

\(=\)3⁰(3+3²+3³)+...+3⁵⁷(3⁵⁸+3⁵⁹+3⁶⁰)

\(=\)3+3²+3³(3⁰+3³+3⁶+...+3⁵⁷)

\(=\)39(3⁰+3³+3⁶+...+3⁵⁷)⋮13

⇒ B⋮13

\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)

Giúp mình cả bài 4,5 ở dưới được ko?

S = (1 / 31 + ... + 1 / 40) + (1 / 41 + ... + 1/ 50) + (1 / 51 + ... + 1 / 60) <
10 / 31 + 10 / 41 + 10 / 51 < 10 / 30 + 10 / 40 + 10 / 50 = 1 / 3 + 1 / 4 + 1 / 5 =
7 / 12 + 1 / 5 < 3 / 5 + 1 / 5 = 4 / 5

=>S<4/5

\(B=3^1+3^2+3^3+...+3^{100}\\3B=3^2+3^3+3^4+...+3^{101}\\3B-B=(3^2+3^3+3^4+...+3^{101})-(3^1+3^2+3^3+...+3^{100})\\2B=3^{101}-3\\\Rightarrow 2B+3=3^{101}\)

Mặt khác: \(2B+3=3^n\)

\(\Rightarrow 3^n=3^{101}\\\Rightarrow n=101(tm)\)

Vậy n = 101.

cảm ơn bạn nha :))