Giải thích các bước giải:
a.Ta có:
$\dfrac ab=\dfrac cd$
$\to \dfrac ab-1=\dfrac cd-1$
$\to\dfrac{a-b}{b}=\dfrac{c-d}d$
2.Ta có:
$\dfrac ab=\dfrac cd$
$\to\dfrac ac=\dfrac bd=\dfrac{2a}{2c}=\dfrac{3b}{3d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}$
$\to\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}$
$\to\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}$
c.Ta có:
$\dfrac ab=\dfrac cd$
$\to (\dfrac ab)^2=(\dfrac cd)^2=\dfrac ab\cdot\dfrac cd$
$\to\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{ab}{cd}=\dfrac{a^2-c^2}{b^2-d^2}$
