Giải thích các bước giải:

$\begin{aligned}
& \text{Đặt } k = \dfrac{a}{b} = \dfrac{c}{d} \text{ (với } k \text{ là hệ số tỉ lệ chung)} \\
& a = bk \\
& c = dk \\
& \text{Chứng minh đẳng thức a:} \\
& \dfrac{2a+b}{3a-b} = \dfrac{2bk+b}{3bk-b} \\
& \dfrac{2a+b}{3a-b} = \dfrac{b(2k+1)}{b(3k-1)} \\
& \dfrac{2a+b}{3a-b} = \dfrac{2k+1}{3k-1} \\
& \dfrac{2c+d}{3c-d} = \dfrac{2dk+d}{3dk-d} \\
& \dfrac{2c+d}{3c-d} = \dfrac{d(2k+1)}{d(3k-1)} \\
& \dfrac{2c+d}{3c-d} = \dfrac{2k+1}{3k-1} \\
& \dfrac{2a+b}{3a-b} = \dfrac{2c+d}{3c-d} \\
& \text{Chứng minh đẳng thức b:} \\
& \dfrac{a^2+b^2}{a^2-b^2} = \dfrac{(bk)^2+b^2}{(bk)^2-b^2} \\
& \dfrac{a^2+b^2}{a^2-b^2} = \dfrac{b^2k^2+b^2}{b^2k^2-b^2} \\
& \dfrac{a^2+b^2}{a^2-b^2} = \dfrac{b^2(k^2+1)}{b^2(k^2-1)} \\
& \dfrac{a^2+b^2}{a^2-b^2} = \dfrac{k^2+1}{k^2-1} \\
& \dfrac{c^2+d^2}{c^2-d^2} = \dfrac{(dk)^2+d^2}{(dk)^2-d^2} \\
& \dfrac{c^2+d^2}{c^2-d^2} = \dfrac{d^2k^2+d^2}{d^2k^2-d^2} \\
& \dfrac{c^2+d^2}{c^2-d^2} = \dfrac{d^2(k^2+1)}{d^2(k^2-1)} \\
& \dfrac{c^2+d^2}{c^2-d^2} = \dfrac{k^2+1}{k^2-1} \\
& \dfrac{a^2+b^2}{a^2-b^2} = \dfrac{c^2+d^2}{c^2-d^2} \\
& \text{Chứng minh đẳng thức c:} \\
& \left(\dfrac{a+b}{c+d}\right)^2 = \left(\dfrac{bk+b}{dk+d}\right)^2 \\
& \left(\dfrac{a+b}{c+d}\right)^2 = \left(\dfrac{b(k+1)}{d(k+1)}\right)^2 \\
& \left(\dfrac{a+b}{c+d}\right)^2 = \left(\dfrac{b}{d}\right)^2 \\
& \left(\dfrac{a+b}{c+d}\right)^2 = \dfrac{b^2}{d^2} \\
& \dfrac{a^2+b^2}{c^2+d^2} = \dfrac{(bk)^2+b^2}{(dk)^2+d^2} \\
& \dfrac{a^2+b^2}{c^2+d^2} = \dfrac{b^2k^2+b^2}{d^2k^2+d^2} \\
& \dfrac{a^2+b^2}{c^2+d^2} = \dfrac{b^2(k^2+1)}{d^2(k^2+1)} \\
& \dfrac{a^2+b^2}{c^2+d^2} = \dfrac{b^2}{d^2} \\
& \left(\dfrac{a+b}{c+d}\right)^2 = \dfrac{a^2+b^2}{c^2+d^2} \\
\end{aligned}$