13
Đặt $\dfrac{a}{b} = \dfrac{c}{d} = k$.
Suy ra $a = bk$ và $c = dk$.
Ta có:
$\dfrac{a^2+ac}{c^2-ac} = \dfrac{(bk)^2+bk \cdot dk}{(dk)^2-bk \cdot dk}$
$= \dfrac{b^2k^2+bdk^2}{d^2k^2-bdk^2}$
$= \dfrac{k^2(b^2+bd)}{k^2(d^2-bd)}$
$= \dfrac{b^2+bd}{d^2-bd}$
Vậy $\dfrac{a^2+ac}{c^2-ac} = \dfrac{b^2+bd}{d^2-bd}$ (ĐPCM)
14,
Từ $\dfrac{a+b}{a-b} = \dfrac{c+d}{c-d}$

`->`
$\dfrac{a+b}{c+d} = \dfrac{a-b}{c-d}$
Áp dụng TCDTSBN cho $\dfrac{a+b}{c+d} = \dfrac{a-b}{c-d}$, ta được:
$\dfrac{a+b+a-b}{c+d+c-d} = \dfrac{a+b-(a-b)}{c+d-(c-d)}$
$\dfrac{2a}{2c} = \dfrac{2b}{2d}$
$\dfrac{a}{c} = \dfrac{b}{d}$
$\dfrac{a}{b} = \dfrac{c}{d}$ (ĐPCM)