\(A=2018\left(20182017-2017\cdot10001\right)+2019\)

=20180000+2019

=20182019

Thiếu dữ kiện, nếu chỉ cho vậy thì không tính đc gt cụ thể của A

+ Làm theo đề là tìm Min của A nhé!

\(A=\frac{a}{2019-c}+\frac{b}{2019-a}+\frac{c}{2019-b}=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}.\)

\(A+3=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)\(\ge\left(a+b+c\right)\frac{9}{2\left(a+b+c\right)}=\frac{9}{2}\)(BĐT Bunhia)

Dấu "=" xra khi a=b=c=2019/3

A \(=\dfrac{2017.20182018}{2018.20172017}\)

ta có : 2017.20182018 = 2017. (2018. 10000 + 2018)

= 2017 . 2018.10000 + 2017 . 2018

2018.20172017 = 2018. (2017. 10000 + 2017)

= 2018 . 2017.10000 + 2018 . 2017

Vậy A \(=\dfrac{2017.20182018}{2018.20172017}\)\(=\dfrac{\text{2017 . 2018.10000 + 2017 . 2018}}{\text{2018 . 2017.10000 + 2018 . 2017}}\)\(=1\)

2018.20172017-2017.20182018=0

a+b=c+d

=>d=a+b-c

Ta có: ab=cd+1

nên: ab-c(a+b-c)=1

=>ab-ac-bc+2c=1

=>a(b-c)-c(b-c)=1

=>(a-c)(b-c)=1

=>a-c=b-c

=>a=b

=>a²⁰¹⁹⁼b²⁰¹⁹

=>a²⁰¹⁹-b²⁰¹⁹=0