$\frac{a}{b} = \frac{c}{d} = 2$のとき、$\frac{a+3c}{b+3d}$の値を求めよ。代数学比例式分数式式の計算2025/7/11. 問題の内容ab=cd=2\frac{a}{b} = \frac{c}{d} = 2ba=dc=2のとき、a+3cb+3d\frac{a+3c}{b+3d}b+3da+3cの値を求めよ。2. 解き方の手順まず、ab=2\frac{a}{b} = 2ba=2とcd=2\frac{c}{d} = 2dc=2より、a=2ba = 2ba=2bc=2dc = 2dc=2dとなることがわかります。次に、a+3cb+3d\frac{a+3c}{b+3d}b+3da+3cにa=2ba = 2ba=2bとc=2dc = 2dc=2dを代入します。a+3cb+3d=2b+3(2d)b+3d\frac{a+3c}{b+3d} = \frac{2b+3(2d)}{b+3d}b+3da+3c=b+3d2b+3(2d)a+3cb+3d=2b+6db+3d\frac{a+3c}{b+3d} = \frac{2b+6d}{b+3d}b+3da+3c=b+3d2b+6da+3cb+3d=2(b+3d)b+3d\frac{a+3c}{b+3d} = \frac{2(b+3d)}{b+3d}b+3da+3c=b+3d2(b+3d)a+3cb+3d=2\frac{a+3c}{b+3d} = 2b+3da+3c=23. 最終的な答え2