Bài 2 :
$a.Đặt\ \dfrac{a}{b}=\dfrac{c}{d}=k$
$⇒a=bk;c=dk$
Ta có :
$\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b(k+1)}{b}=k+1$
$\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d(k+1)}{b}=k+1$
$⇒\dfrac{a+b}{b}=\dfrac{c+d}{d}$
b.Ta có :
$\dfrac{a}{b}=\dfrac{c}{d}⇒\dfrac{a}{c}=\dfrac{b}{d}$
Áp dụng tính chất của dãy tỉ số bằng nhau :
$⇒\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a}{3c}=\dfrac{5b}{3d}=\dfrac{5a+5b}{3c+3d}$
$\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a}{3c}=\dfrac{5b}{3d}=\dfrac{5a-5b}{3c-3d}$
$⇒\dfrac{5a+5b}{3c+3d}=\dfrac{5a-5b}{3c-3d}$
Bài 3 :
$a.ab.bc.ca=\dfrac{3}{5}.\dfrac{4}{5}.\dfrac{3}{4}$
$⇔(abc)^2=\dfrac{9}{25}$
$⇒abc=±\dfrac{3}{25}$
⇒\(\left[ \begin{array}{l}abc=\dfrac{3}{5}\\abc=\dfrac{-3}{5}\end{array} \right.\) ⇔\(\left[ \begin{array}{l}\left\{\begin{matrix} a=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4} & \\ b=\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{4}{5} & \\ c=\dfrac{3}{5}:\dfrac{3}{5}=1 & \end{matrix}\right.\\\left\{\begin{matrix} a=\dfrac{-3}{5}:\dfrac{4}{5}=\dfrac{-3}{4} & \\ b=\dfrac{-3}{5}:\dfrac{3}{4}=\dfrac{-4}{5} & \\ c=\dfrac{-3}{5}:\dfrac{3}{5}=-1 & \end{matrix}\right.\end{array} \right.\)
