Giải thích các bước giải:
ĐK: $b+d\ne 0; ad-bc\ne 0$
Ta có:
$\begin{array}{l}
\dfrac{{b + c}}{{b + d}} - \dfrac{{ac - bd}}{{ad - bc}}\\
= \dfrac{{\left( {b + c} \right)\left( {ad - bc} \right) - \left( {ac - bd} \right)\left( {b + d} \right)}}{{\left( {b + d} \right)\left( {ad - bc} \right)}}\\
= \dfrac{{abd + acd - {b^2}c - b{c^2} - abc - acd + {b^2}d + b{d^2}}}{{\left( {b + d} \right)\left( {ad - bc} \right)}}\\
= \dfrac{{abd - abc - {b^2}c - b{c^2} + {b^2}d + b{d^2}}}{{\left( {b + d} \right)\left( {ad - bc} \right)}}\\
= \dfrac{{\left( {abd + {b^2}d + b{d^2}} \right) - \left( {abc + {b^2}c + b{c^2}} \right)}}{{\left( {b + d} \right)\left( {ad - bc} \right)}}\\
= \dfrac{{bd\left( {a + b + d} \right) - bc\left( {a + b + c} \right)}}{{\left( {b + d} \right)\left( {ad - bc} \right)}}\\
= \dfrac{{bd.\left( { - c} \right) - bc\left( { - d} \right)}}{{\left( {b + d} \right)\left( {ad - bc} \right)}}\\
= \dfrac{0}{{\left( {b + d} \right)\left( {ad - bc} \right)}}\\
= 0
\end{array}$
$ \Rightarrow \dfrac{{b + c}}{{b + d}} = \dfrac{{ac - bd}}{{ad - bc}}$
$\to $ Điều phải cm.
