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驕カ螂�スソ�ス髯橸スサ隴会スョ陝キ謌頑€ヲ�ス�ャ髯滓汚�ソ�ス(x+a)(x+b)
驕カ螂�スソ�ス髯キ�キ�ス�ス(2)
驕カ螂�スソ�ス髯キ�キ�ス�ス(3)
驕カ螂�スソ�ス髯橸スサ隴会スョ陝キ謌頑€ヲ�ス�ャ髯滓汚�ソ�ス(a+b)2
驕カ螂�スソ�ス髯キ�キ�ス�ス(2)
驕カ螂�スソ�ス髯キ�キ�ス�ス(3)
驕カ螂�スソ�ス髯キ�キ�ス�ス(4)
驕カ螂�スソ�ス髯橸スサ隴会スョ陝キ謌頑€ヲ�ス�ャ髯滓汚�ソ�ス(a+b)(a−b)
驕カ螂�スソ�ス髯キ�キ�ス�ス(2)
驕カ螂�スソ�ス髯キ�キ�ス�ス(3)-髴托スエ�ス�セ髯懶スィ�ス�ィ髯懶スィ�ス�ー
驕カ螂�スソ�ス髯橸スサ隴会スョ陝キ謌頑€ヲ�ス�ャ髯滓汚�ソ�ス(驍オ�コ�ス�セ驍オ�コ�ス�ィ驛「�ァ�ス�ス)
驕カ螂�スソ�ス髯橸スサ隴会スョ陝キ謌頑€ヲ�ス�ャ髯滓汚�ソ�ス(驍オ�コ�ス�セ驍オ�コ�ス�ィ驛「�ァ�ス�ス,髯懶ソス陜楢カ」�ス�。�ス�ス)
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【解説】(問題は下にあります.)
○書く順序だけを変えても、式の値は変わりませんので
(−a+b)=(b−a)・・・(1)です.
○引き算の順序を変えると符号が逆になりますので  
(b−a)=−(a−b)・・・(2)です.

○これらを使って、次のような展開ができます.
(例1)
(p+q)(−p+q)
       ← (1)を使う
=(p+q)(q−p)
       ←  p+q=q+p を使う
=(q+p)(q−p)
       ←  (a+b)(a−b)=a2−b2 を使う
=q2−p2 ・・・(答)

※方法は一つとは限りません.(例1と同じ式について別の解き方)
(p+q)(−p+q)
       ← (2)を使う
=−(p+q)(p−q)
       ←  (a+b)(a−b)=a2−b2 を使う
=−(p2−q2)
       ← (2)を使う
=q2−p2   ・・・(答)
注意
のように、とにかくバラバラにしてもできますが、その練習だけでは、元に戻す(=次に出てくる因数分解)見通しが立たなくなります。
ここでは、なるべく公式を用いて展開する練習をします。

【問題】 次の式を展開しなさい.(正しい式をマウスで選びなさい.)
(1) (3+2x)(2x−3)

=

(2) (2−y)(y+2)

=

(3) (−x−7)(−x+7)

=

(4) (−4a+3b)(4a+3b)

=

(5) (−3a+5b)(−3a−5b)

=

(6)

=

(7) (2x−3y)(−2x−3y)

=

(8)

=


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■[個別の頁からの質問に対する回答][展開公式について/17.3.7]
中3 式の展開 展開公式(a+b)(a-b) ルートの計算はまだ習っていないです
=>[作者]:連絡ありがとう.最後の一題はルートの計算を習ってからやってください.

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