✨ Best Answer ✨
(ac+bd)=X, (ad+bc)=Y とする
X²-Y²=(X+Y)(X-Y)
よって
{ac+bd+(ad+bc)}{ac+bd-(ad+bc)}
=(ac+ad+bc+bd)(ac-ad-bc+bd)
={a(c+d)+b(c+d)}{a(c-d)-b(c-d)}
=(a+b)(c+d)(a-b)(c-d)
=(a+b)(a-b)(c+d)(c-d)
✨ Best Answer ✨
(ac+bd)=X, (ad+bc)=Y とする
X²-Y²=(X+Y)(X-Y)
よって
{ac+bd+(ad+bc)}{ac+bd-(ad+bc)}
=(ac+ad+bc+bd)(ac-ad-bc+bd)
={a(c+d)+b(c+d)}{a(c-d)-b(c-d)}
=(a+b)(c+d)(a-b)(c-d)
=(a+b)(a-b)(c+d)(c-d)
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ありがとうございました!