TH1: a < b

=> 2019a < 2019b

=> ab + 2019a < ab+ 2019b

=> a(b+2019) < b(a+2019)

=> a/b < (a+2019)/(b+2019)

TH2: a = b

=> a/b = (a+2019)/(b+2019)

TH3: a > b

=> ab + 2019a > ab+ 2019b

=> a(b+2019) > b(a+2019)

=> a/b > (a+2019)/(b+2019)

đúng ko moonshine

đầu tiên: a < b

=> 2019a < 2019b

=> ab + 2019a < ab+ 2019b

=> a(b+2019) < b(a+2019)

=> a/b < (a+2019)/(b+2019)

2: a = b

=> a/b = (a+2019)/(b+2019)

3: a > b

=> ab + 2019a > ab+ 2019b

=> a(b+2019) > b(a+2019)

=> a/b > (a+2019)/(b+2019)

TH1: a < b

=> 2019a < 2019b

=> ab + 2019a < ab + 2019b

=> a (b + 2019) < b (a + 2019)

=> \(\frac{a}{b}\) < \(\frac{a+2019}{b+2019}\).

TH2: a = b

=> \(\frac{a}{b}\) = \(\frac{a+2019}{b+2019}\).

TH3: a > b

=> ab + 2019a > ab + 2019b

=> a (b + 2019) > b (a + 2019)

=> \(\frac{a}{b}\) > \(\frac{a+2019}{b+2019}\).

Chúc bạn học tốt!

Có \(a\left(b+1\right)< b\left(a+1\right)\Leftrightarrow ab+a< ab+b\)

\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)

Áp dụng \(\frac{2^{2018}}{3^{2019}}< \frac{2^{2018}+1}{3^{2019}+1}\)

Ta có:

\(1-\frac{a}{b}=\frac{b-a}{b}\)

\(1-\frac{a+1}{b+1}=\frac{b+1-a-1}{b+1}=\frac{b-a}{b+1}\)

Vì b < b + 1 và a < b; a, b nguyên dương  => b - a > 0 nên \(\frac{b-a}{b}>\frac{b-a}{b+1}\)

Do đó \(1-\frac{a}{b}>1-\frac{a+1}{b+1}\)

\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)

Áp dụng chứng minh tương tự nhé bạn

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\) 

Với  :   \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\) 

Và   :   \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\) 

             \(B=1-\frac{1}{2020}< 1< A\)

\(A=\frac{7^{2018}+1}{7^{2019}+1}\)

\(\Rightarrow7A=\frac{7^{2019}+7}{7^{2019}+1}=1+\frac{6}{7^{2019}+1}\)

\(B=\frac{7^{2019}+1}{7^{2020}+1}\)

\(\Rightarrow7B=\frac{7^{2020}+7}{7^{2020}+1}\)

\(\Rightarrow7B=1+\frac{6}{7^{2020}+1}\)

Vì 7 ^ 2019 < 7 ^ 2020 => 7 ^ 2019 + 1 < 7 ^ 2020 + 1

=> 6 / ( 7 ^ 2019 + 1 ) > 6 / ( 7 ^ 2020 + 1 )  

=> 1 + 6 / ( 7 ^ 2019 + 1 ) > 1 + 6 / ( 7 ^ 2020 + 1 )  

=> 7A > 7B

Vì A , B > 0 

Nên A > B 

Vì \(7^{2018}< 7^{2019}\)nên \(7^{2018}+1< 7^{2019}+1\)

\(\Rightarrow\frac{7^{2018}+1}{7^{2019}+1}< \frac{7^{2019}+1}{7^{2019}+1}\)

Hay A < B

Chúc bạn học tốt ! Nguyễn Thi An Na

B>A

nha bn

vì sao lại lớn hơn

Ta có thể thấy rõ: \(2019< 10^{2019};2019< 10^{2018}\)

\(\Rightarrow-\frac{9}{2019}>-\frac{9}{10^{2019}};-\frac{19}{2019}>-\frac{19}{10^{2018}}\)(do tử là số âm nên mẫu càng lớn thì cả pso càng nhỏ)

\(\Rightarrow-\frac{9}{2019}+\frac{-19}{2019}>-\frac{9}{10^{2019}}+\frac{-19}{10^{2018}}\)

Hay A>B

Chúc bn hok tốt,nhớ k nha

Vì b > 0 => b + 2019 > 0

Ta có: \(\frac{a}{b}=\frac{a.\left(b+2019\right)}{b.\left(b+2019\right)}=\frac{a.b+a.2019}{b.\left(b+2019\right)}=\frac{a+2019}{b+2019}=\)

\(\frac{b.\left(a+2019\right)}{b.\left(b+2019\right)}=\frac{a.b+b.2019}{b.\left(b+2019\right)}\)

TH1: Nếu a < b => \(\frac{a.b+a.2019}{b.\left(b+2019\right)}< \frac{a.b+b.2019}{b.\left(b+2019\right)}\)

                       hay \(\frac{a}{b}< \frac{a+2019}{b+2019}\)

TH2: Nếu a = b => \(\frac{a.b+a.2019}{b.\left(b+2019\right)}=\frac{a.b+b.2019}{b.\left(b+2019\right)}\)

                       hay \(\frac{a}{b}=\frac{a+2019}{b+2019}\)

TH3: Nếu a > b => \(\frac{a.b+a.2019}{b.\left(b+2019\right)}>\frac{a.b+b.2019}{b.\left(b+2019\right)}\)

                       hay \(\frac{a}{b}=\frac{a+2019}{b+2019}\)

Xét tích : \(a(b+2019)=ab+2019a\)

\(b(a+2019)=ab+2019b\)

Vì b > 0 nên b + 2019 > 0

Nếu a > b thì \(ab+2019a>ab+2019b\)

\(a(b+2019)>b(a+2019)\)

\(\Rightarrow\frac{a}{b}>\frac{a+2019}{b+2019}\)

Nếu a < b thì \(ab+2019a< ab+2019b\)

\(a(b+2019)< b(a+2019)\)

\(\Rightarrow\frac{a}{b}< \frac{a+2019}{b+2019}\)

Nếu a = b thì rõ ràng \(\frac{a}{b}=\frac{a+2019}{b+2019}\)