与えられた式 $ax^2y^2 - ax^2 - ay^2 + a$ を因数分解します。代数学因数分解多項式式の展開2025/4/161. 問題の内容与えられた式 ax2y2−ax2−ay2+aax^2y^2 - ax^2 - ay^2 + aax2y2−ax2−ay2+a を因数分解します。2. 解き方の手順まず、最初の2項と最後の2項をそれぞれ aaa でくくります。ax2y2−ax2−ay2+a=a(x2y2−x2)−a(y2−1)ax^2y^2 - ax^2 - ay^2 + a = a(x^2y^2 - x^2) - a(y^2 - 1)ax2y2−ax2−ay2+a=a(x2y2−x2)−a(y2−1)次に、最初の aaa でくくられた項を x2x^2x2 でくくり、2番目の aaa でくくられた項をそのままにします。a(x2y2−x2)−a(y2−1)=ax2(y2−1)−a(y2−1)a(x^2y^2 - x^2) - a(y^2 - 1) = a x^2 (y^2 - 1) - a(y^2 - 1)a(x2y2−x2)−a(y2−1)=ax2(y2−1)−a(y2−1)次に、全体を a(y2−1)a(y^2-1)a(y2−1) でくくります。ax2(y2−1)−a(y2−1)=a(y2−1)(x2−1)a x^2 (y^2 - 1) - a(y^2 - 1) = a (y^2 - 1) (x^2 - 1)ax2(y2−1)−a(y2−1)=a(y2−1)(x2−1)最後に、y2−1y^2-1y2−1 と x2−1x^2-1x2−1 をそれぞれ (y−1)(y+1)(y-1)(y+1)(y−1)(y+1) と (x−1)(x+1)(x-1)(x+1)(x−1)(x+1) に因数分解します。a(y2−1)(x2−1)=a(y−1)(y+1)(x−1)(x+1)a (y^2 - 1) (x^2 - 1) = a(y-1)(y+1)(x-1)(x+1)a(y2−1)(x2−1)=a(y−1)(y+1)(x−1)(x+1)3. 最終的な答えa(x−1)(x+1)(y−1)(y+1)a(x-1)(x+1)(y-1)(y+1)a(x−1)(x+1)(y−1)(y+1)