CMR: Với (a + b + c + d).(a – b – c + d) = (a – b + c – d).(a + b – c – d) thì $\frac{a}{c}$ = $\frac{b}{d}$

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1. (a+b+c+d)(ab−c+d)=(a−b+c−d)(a+b−c−d)

   [(a+d)+(b+c)][(a+d)−(b+c)]=[(a−d)−(b−c)][(a−d)+(b−c)]⇔(a+d)2−(b+c)2=(a−d)2−(b−c)2

   ⇔a2+2ad+d2−b2−2bc−c2=a2−2ad+d2−b2+2bc−c2

   ⇔2ad−2bc=−2ad+2bc⇔4ad=4bc⇔ad=bc⇔ac=bd(đpcm).

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2. \(\begin{array}{l}
   \left( {a + b + c + d} \right)\left( {a – b – c + d} \right) = \left( {a – b + c – d} \right)\left( {a + b – c – d} \right)\\
    \Leftrightarrow \left[ {\left( {a + d} \right) + \left( {b + c} \right)} \right]\left[ {\left( {a + d} \right) – \left( {b + c} \right)} \right] = \left[ {\left( {a – d} \right) – \left( {b – c} \right)} \right]\left[ {\left( {a – d} \right) + \left( {b – c} \right)} \right]\\
    \Leftrightarrow {\left( {a + d} \right)^2} – {\left( {b + c} \right)^2} = {\left( {a – d} \right)^2} – {\left( {b – c} \right)^2}\\
    \Leftrightarrow {a^2} + 2ad + {d^2} – {b^2} – 2bc – {c^2} = {a^2} – 2ad + {d^2} – {b^2} + 2bc – {c^2}\\
    \Leftrightarrow 2ad – 2bc =  – 2ad + 2bc\\
    \Leftrightarrow 4ad = 4bc\\
    \Leftrightarrow ad = bc\\
    \Leftrightarrow \frac{a}{c} = \frac{b}{d}\,\,\,\left( {dpcm} \right).
   \end{array}\)

    

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