Đáp án:
Giải thích các bước giải:
$\dfrac{a+c}{b+d}-\dfrac{a}{b}$
$=\dfrac{(a+c)b}{b(b+d)}-\dfrac{a(b+d)}{b(b+d)}$
$=\dfrac{ab+bc}{b(b+d)}-\dfrac{ab+ad}{b(b+d)}$
$=\dfrac{ab+bc-ab-ad}{b(b+d)}$
$=\dfrac{bc-ad}{b(b+d)}>0$
$⇒\dfrac{a+c}{b+d}-\dfrac{a}{b}>0$
$⇒\dfrac{a+c}{b+d}>\dfrac{a}{b} (1)$
$\dfrac{c}{d}-\dfrac{a+c}{b+d}$
$=\dfrac{c(b+d)}{d(b+d)}-\dfrac{(a+c)d}{d(b+d)}$
$=\dfrac{bc+cd-ad-cd}{d(b+d)}$
$=\dfrac{bc-ad}{d(b+d)} >0$
$⇒\dfrac{c}{d}-\dfrac{a+c}{b+d}>0$
$⇒\dfrac{c}{d}>\dfrac{a+c}{b+d}$ $(2)$
Từ $(1)$ và $(2) ⇒ \dfrac{a}{b}<\dfrac{a+c}{b+d}<\dfrac{c}{d}$