A là số dương, B là số âm => A>B

A có 50 thừa số âm

=> A > 0

b) CÓ 49 thừa số âm 

=> B < 0 

giup toi cai

A = 1 + 4 + 4^2 + ... + 4^99

4A = 4 + 4^2 + 4^3 +... + 4^100

4A - A = 3A = ( 4 + 4^2 + 4^100 ) - ( 1 + 4 + 4^2 + 4^99 )

3A = 4^100 - 1

Ta thấy: 3A < B => A < B/3 ( đpcm )

k đúng nhé

A = 99^2015 + 1/99^2014 + 1 < 99^2015 + 1 + 98 / 99^2014 + 1 + 98

                                                  = 99^2015 + 99 / 99^2014 + 99

                                                  = 99(99^2014 + 1) / 99(99^2013+1)

                                                  = 99^2014 + 1 / 99^2013 + 1 = B

=> A < B

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

1/51+1/52+....+1/99<1/2

Ta có: \(\frac{A}{2020}=\frac{2020^{100}-1}{2020^{100}+2020}=\frac{2020^{100}+2020-2021}{2020^{100}+2020}=1-\frac{2021}{2020^{100}+2020}\)

\(\frac{B}{2020}=\frac{2020^{101}-1}{2020^{101}+2020}=\frac{2020^{101}+2020-2021}{2020^{101}+2020}=1-\frac{2021}{2020^{101}+2020}\)

Ta có: \(2020^{100}+2020<2020^{101}+2020\)

=>\(\frac{2021}{2020^{100}+2020}>\frac{2021}{2020^{101}+2020}\)

=>\(-\frac{2021}{2020^{100}+2020}<-\frac{2021}{2020^{101}+2020}\)

=>\(-\frac{2021}{2020^{100}+2020}+1<-\frac{2021}{2020^{101}+2020}+1\)

=>\(\frac{A}{2020}<\frac{B}{2020}\)

=>A<B

Chắc chắn (x-1)+2 > (x-1)