$(x+\frac{1}{3})(x+\frac{2}{3})$ を展開せよ。代数学展開多項式2025/5/171. 問題の内容(x+13)(x+23)(x+\frac{1}{3})(x+\frac{2}{3})(x+31)(x+32) を展開せよ。2. 解き方の手順与えられた式を展開します。(x+13)(x+23)=x(x+23)+13(x+23)(x+\frac{1}{3})(x+\frac{2}{3}) = x(x+\frac{2}{3}) + \frac{1}{3}(x+\frac{2}{3})(x+31)(x+32)=x(x+32)+31(x+32)=x2+23x+13x+13⋅23= x^2 + \frac{2}{3}x + \frac{1}{3}x + \frac{1}{3} \cdot \frac{2}{3}=x2+32x+31x+31⋅32=x2+(23+13)x+29= x^2 + (\frac{2}{3}+\frac{1}{3})x + \frac{2}{9}=x2+(32+31)x+92=x2+33x+29= x^2 + \frac{3}{3}x + \frac{2}{9}=x2+33x+92=x2+x+29= x^2 + x + \frac{2}{9}=x2+x+923. 最終的な答えx2+x+29x^2 + x + \frac{2}{9}x2+x+92