Giải thích các bước giải:

a.Ta có:

$\dfrac ab=\dfrac cd$

$\to \dfrac ac=\dfrac bd=\dfrac{3a+5b}{3c+5d}=\dfrac{6a-7b}{6c-7d}$

$\to \dfrac{3a+5b}{6a-7b}=\dfrac{3c+5d}{6c-7d}$

b.Ta có: 

$\dfrac ab=\dfrac cd=\dfrac{2a}{2b}=\dfrac{3c}{3d}=\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}$

c.Đề sai

d.$\dfrac{a}b=\dfrac{c}d$

$\to \dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}$