Ta có:

\(\dfrac{1}{20^2}< \dfrac{1}{20\cdot19}=\dfrac{1}{19}-\dfrac{1}{20}\)

\(\dfrac{1}{21^2}< \dfrac{1}{20\cdot21}=\dfrac{1}{20}-\dfrac{1}{21}\)

\(...\)

\(\dfrac{1}{30^2}< \dfrac{1}{29\cdot30}=\dfrac{1}{29}-\dfrac{1}{30}\)

\(\Rightarrow A< \dfrac{1}{19}-\dfrac{1}{30}< \dfrac{1}{19}\)

(1/2)¹⁵⁰⁰=(1/2⁶)²⁵⁰=(1/64)²⁵⁰

Do 1/16>1/64 =>(1/16)²⁵⁰>(1/64)²⁵⁰

Vậy (1/16)²⁵⁰>(1/2)¹⁵⁰⁰

\(\left(\frac{1}{16}\right)^{250}\) và \(\left(\frac{1}{2}\right)^{1500}\)

=> \(\left(\frac{1}{16}\right)^{250}\) và \(\left(\left(\frac{1}{2}\right)^6\right)^{250}\)

=> \(\frac{1}{16}\) và \(\left(\frac{1}{2}\right)^6\)

=> \(\frac{1}{16}\) và \(\frac{1}{64}\)

=>  \(\frac{1}{16}\) >  \(\frac{1}{64}\)  hay  \(\left(\frac{1}{16}\right)^{250}\) >  \(\left(\frac{1}{2}\right)^{1500}\)

HD ; em đổi A và B thành các Thừa số Nguyên tố rồi so sánh

Ta có: A= 2^46.125^7=2^22.2^26.5^35        (1)

B=9^11.6^22=3^44.2^22           (2)

Từ (1)(2)=> A>B

\(A>B\)nhé 

hok tốt/skr

Nêu rõ cách làm

12/13>22/33

tích mình mình tích lại cho 

Lưu ý : phải gửi tin nhắn

\(\frac{12}{13}>\frac{22}{33}\)

OK

12/13<22/33

22³³ < 33²²

Học tốt!!!

\(22^{33}>33^{22}\)

Ta có:

 \(22^{33}=\left(2.11\right)^{33}=2^{33}.11^{33}=2^{3^{11}}.11^{22}.11^{11}=8^{11}.11^{22}.11^{11}\)

\(33^{22}=\left(3.11\right)^{22}=3^{22}.11^{22}=3^{2^{11}}.11^{22}=9^{11}.11^{22}\)

Ta thấy: \(11^{11}>9^{11}\)

Vậy \(22^{33}>33^{22}\)

Hok tốt ~~~

5 mũ 23 bé hơn

a) \(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)

b) \(B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+...+3^{101}\)

\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)

\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)

c) \(C=5+5^2+...+5^{30}\)

\(\Rightarrow5C=5^2+5^3+...+5^{31}\)

\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)

\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)

d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)

\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)

\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)