Cho a,b,c \(\in\)N* và a<b<1.Ta có:\(\frac{a}{b}<\frac{a+c}{b+c}\)

\(\Rightarrow\)a(b+c)<b(a+c)

\(\Rightarrow\)ab+ac<ba+bc

\(\Rightarrow\)ac<bc

Tiếp nè:

\(\Rightarrow\)a<b đúng

Mặt khác:\(\frac{1}{2}<\frac{1+1}{2+1}=\frac{2}{3}\)

              \(\frac{3}{4}<\frac{3+1}{4+1}=\frac{4}{5}\)

               \(\frac{199}{200}<\frac{199+1}{200+1}=\frac{200}{201}\)

\(\Rightarrow A<\frac{2}{3}.\frac{4}{5}...........\frac{200}{201}\)

\(\Rightarrow A^2<\frac{1}{2}.\frac{2}{3}.\frac{3}{4}............\frac{199}{200}.\frac{200}{201}\)

\(\Rightarrow A^2<\frac{1}{101}<\frac{1}{100}\)

\(\Rightarrow A<\frac{1}{10}\)

b,Chưa làm được,sorry

ko pit

>

chắc chắn

199²⁰ =( (200 -1)²)¹⁰ =( 200² - 400 +1)¹⁰ = 1601¹⁰  > 200²⁰

=> 199²⁰ > 200¹⁵

A=\(\frac{1}{100}\)+\(\frac{1}{101}\)+\(\frac{1}{102}\)+...+\(\frac{1}{200}\)

   (Sử dung phương pháp chặn số đầu)

\(\frac{1}{100}\)>\(\frac{1}{101}\)

\(\frac{1}{100}\)>\(\frac{1}{102}\)

           ...

\(\frac{1}{100}\)>\(\frac{1}{200}\)

nên \(\frac{1}{100}\)+\(\frac{1}{101}\)+\(\frac{1}{102}\)+...+\(\frac{1}{200}\)> \(\frac{1}{100}\)+\(\frac{1}{100}\)+...+\(\frac{1}{100}\)(có 101 phân số)

\(\Rightarrow\)\(\frac{1}{100}\)+\(\frac{1}{101}\)+\(\frac{1}{102}\)+...+\(\frac{1}{200}\)>101.\(\frac{1}{100}\)=\(\frac{101}{100}\)>1>\(\frac{3}{4}\)

\(\Rightarrow\)A >\(\frac{3}{4}\)

a, 9^5>27^3

​b,3^200>2^300

​c, 32^11<17^14

a) A = 4^200 - 4

b) 3A = 4^200 . 3 - 12 > 4^200

c) nghĩ đã   

b) >

Tính:

A= 1.3+3.5+5.7+...+99.101

B= 1^2+3^2+5^2+...+99^2