a: Ta có

A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)

⇒ A > \(\dfrac{1}{10}\) + \((\dfrac{1}{100}\) + \(\dfrac{1}{100}\) + ...+ \(\dfrac{1}{100}\)\()\)90 số hạng 

⇒ A > \(\dfrac{1}{10}\) + \(\dfrac{90}{100}\)

⇒ A > 1

vậy A > 1

b: ta có

S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))

⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))

⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)

⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)

⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)

vậy S > \(\dfrac{1}{2}\)

toán lớp 2

bt ko mà nói ^^

tui ghét lũy thừa nên lười

mik cx ko bt câu này

mik cx dg định đăng câu này

hok tốt

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) \(S=1+4+4^2+4^3+...4^{62}\)

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

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

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

\(\Rightarrow dpcm\)

Ta có : \(S=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{199}+\frac{1}{200}\)

\(\Rightarrow S>\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\) ( 181 phân số )

\(\Rightarrow S>\frac{181}{200}>\frac{180}{200}=\frac{9}{10}\)

\(\Rightarrow S>\frac{9}{10}\)       \(\Rightarrowđpcm\)

C = 120120 + 121121 + 122122 + ... + 12001200

⇒ CC> 12001200 + 12001200 + 12001200 + ...... + 12001200 ( 181181 phân số )

⇒ CC > 181200181200 > 180200180200 = 910910

⇒ CC >910

cs chép sai đè ko vậy

không

tự giải đi

S = \(\frac{1}{20}+\frac{1}{21}...+\frac{1}{199}+\frac{1}{200}\)  ( có 181 phân số )

=> S > \(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}+\frac{1}{200}\)

=> S > \(\frac{1}{200}.181\)

=> S > \(\frac{181}{200}\)> \(\frac{180}{200}\)= \(\frac{9}{10}\)

Vậy S > 9 / 10

GIÚP NHA , AI LÀM ĐƯƠC 1 NGÀY TK 3TK

S = \(\frac{1}{20}\)+ \(\frac{1}{21}\)+ ....+\(\frac{1}{200}\)có 181 p/s

mà \(\frac{1}{20}\)>\(\frac{1}{200}\)

.............

    \(\frac{1}{199}\)>\(\frac{1}{200}\)

    \(\frac{1}{200}\)=\(\frac{1}{200}\)

nên  ta có     S > \(\frac{1}{200}\)+ \(\frac{1}{200}\)+..... có 181 phân số \(\frac{1}{200}\)

vậy \(\frac{1}{200}\)*181=\(\frac{181}{200}\)mà \(\frac{181}{200}\)>\(\frac{9}{10}\)mà \(\frac{1}{20}\)+......+\(\frac{1}{200}\)(có 181 số)>\(\frac{1}{200}\)+\(\frac{1}{200}\)(có 181 p/s \(\frac{1}{200}\))>\(\frac{9}{10}\)

Vậy ==> S>\(\frac{9}{10}\)