Giải thích các bước giải:
\[\begin{array}{l}
dat:\frac{a}{b} = \frac{c}{d} = k = > a = kb;c = kd\\
A = \frac{{{a^2} + ac}}{{{c^2} - ac}} = \frac{{{b^2}{k^2} + {k^2}bd}}{{{k^2}{d^2} - {k^2}bd}} = \frac{{b{k^2}(b + d)}}{{d{k^2}(d - b)}} = \frac{{b(b + d)}}{{d(d - b)}}\\
B = \frac{{{b^2} + bd}}{{{d^2} - bd}} = \frac{{b(b + d)}}{{d(d - b)}}\\
= > A = B = > \frac{{{a^2} + ac}}{{{b^2} + bd}} = \frac{{{c^2} - ac}}{{{d^2} - bd}}(dpcm)
\end{array}\]