Giúp mình với!!!! `\text{Cho abc=2022.Rút gọn:}` `\text{A=}`$\frac{a}{ab+a+2022}+$ $\frac{b}{bc+b+1}+$ $\frac{2022}{ac+2022c+2022}$

Đáp án:

Giải thích các bước giải:

$\  A = \dfrac{a}{ab + a + 2022} + \dfrac{b}{bc + b + 1} + \dfrac{2022c}{ac + 2022c + 2022} $

$\ ⇒ A = \dfrac{abc}{(ab + a + 2022)bc} + \dfrac{b}{bc + b + 1} + \dfrac{2022bc}{(ac + 2022c + 2022)b} $

$\ ⇒ A = \dfrac{abc}{abbc + abc + 2022bc} + \dfrac{b}{bc + b + 1} + \dfrac{2022bc}{acb + 2022cb + 2022b} $

Mà $\ abc = 2022 $

$\ ⇒ A = \dfrac{2022}{2022b + 2022 + 2022bc} + \dfrac{b}{bc + b + 1} + \dfrac{2022bc}{2022 + 2022cb + 2022b} $

$\ ⇒ A = \dfrac{2022}{2022(b + 1 + bc)} + \dfrac{b}{bc + b + 1} + \dfrac{2022bc}{2022(1 + cb + b)} $

$\ ⇒ A= \dfrac{1}{b + 1 + bc} + \dfrac{b}{bc + b + 1} + \dfrac{bc}{1 + cb + b} $

$\ ⇒ A = \dfrac{1 + b + bc}{bc + b + 1} $

$\ ⇒ A = 1 $

Vậy $\ A = 1 $