$\frac{\sqrt{3}}{\sqrt{54}} + \frac{5}{3\sqrt{2}}$ を計算して簡単にせよ。算数平方根有理化計算2025/6/301. 問題の内容354+532\frac{\sqrt{3}}{\sqrt{54}} + \frac{5}{3\sqrt{2}}543+325 を計算して簡単にせよ。2. 解き方の手順まず、54\sqrt{54}54 を簡単にします。54=9×6=9×6=36\sqrt{54} = \sqrt{9 \times 6} = \sqrt{9} \times \sqrt{6} = 3\sqrt{6}54=9×6=9×6=36したがって、354=336\frac{\sqrt{3}}{\sqrt{54}} = \frac{\sqrt{3}}{3\sqrt{6}}543=363 です。次に、336\frac{\sqrt{3}}{3\sqrt{6}}363 の分母を有理化します。336=336×66=183×6=9×218=3218=26\frac{\sqrt{3}}{3\sqrt{6}} = \frac{\sqrt{3}}{3\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{18}}{3 \times 6} = \frac{\sqrt{9 \times 2}}{18} = \frac{3\sqrt{2}}{18} = \frac{\sqrt{2}}{6}363=363×66=3×618=189×2=1832=62次に、532\frac{5}{3\sqrt{2}}325 の分母を有理化します。532=532×22=523×2=526\frac{5}{3\sqrt{2}} = \frac{5}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{2}}{3 \times 2} = \frac{5\sqrt{2}}{6}325=325×22=3×252=652したがって、354+532=26+526=2+526=626=2\frac{\sqrt{3}}{\sqrt{54}} + \frac{5}{3\sqrt{2}} = \frac{\sqrt{2}}{6} + \frac{5\sqrt{2}}{6} = \frac{\sqrt{2} + 5\sqrt{2}}{6} = \frac{6\sqrt{2}}{6} = \sqrt{2}543+325=62+652=62+52=662=23. 最終的な答え2\sqrt{2}2