Giải thích các bước giải:
$a)\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}$
$⇒(a+b)(c-a)=(a-b)(c+a)$
$⇒ac+bc-a^2-ba=ac-bc+a^2-ba$
$⇒-bc+a^2=-a^2+bc$
$⇒2a^2=2bc$
$⇒a^2=bc$
$⇒aa=bc$
$⇒\dfrac{a}{b}=\dfrac{c}{a}\text{(đpcm)}$
$b)A=|x-2020|+|x-2021|$
$=|x-2020|+|2021-x|$
Áp dụng BĐT: $|A+B|≤|A|+|B|$
$⇒|x-2020|+|2021-x|≥|x-2020+2021-x|$
$⇒|x-2020|+|2021-x|≥1$
$⇒|x-2020|+|x-2021|≥1$
Vậy $A_{min}=1$ tại $2020≤x≤2021$
