次の式を満たす $\boxed{}$ に当てはまる数を求める問題です。 $1 \frac{2}{3} \div 5.5 + 1 \frac{7}{12} \div (2.25 - \boxed{}) = 1 \frac{1}{6}$算数分数四則演算方程式2025/4/151. 問題の内容次の式を満たす \boxed{} に当てはまる数を求める問題です。123÷5.5+1712÷(2.25−)=1161 \frac{2}{3} \div 5.5 + 1 \frac{7}{12} \div (2.25 - \boxed{}) = 1 \frac{1}{6}132÷5.5+1127÷(2.25−)=1612. 解き方の手順まず、与えられた式を整理します。123=531 \frac{2}{3} = \frac{5}{3}132=355.5=1125.5 = \frac{11}{2}5.5=2111712=19121 \frac{7}{12} = \frac{19}{12}1127=1219116=761 \frac{1}{6} = \frac{7}{6}161=672.25=942.25 = \frac{9}{4}2.25=49与えられた式に代入すると、53÷112+1912÷(94−)=76\frac{5}{3} \div \frac{11}{2} + \frac{19}{12} \div (\frac{9}{4} - \boxed{}) = \frac{7}{6}35÷211+1219÷(49−)=6753×211+1912÷(94−)=76\frac{5}{3} \times \frac{2}{11} + \frac{19}{12} \div (\frac{9}{4} - \boxed{}) = \frac{7}{6}35×112+1219÷(49−)=671033+1912÷(94−)=76\frac{10}{33} + \frac{19}{12} \div (\frac{9}{4} - \boxed{}) = \frac{7}{6}3310+1219÷(49−)=671912÷(94−)=76−1033\frac{19}{12} \div (\frac{9}{4} - \boxed{}) = \frac{7}{6} - \frac{10}{33}1219÷(49−)=67−3310右辺を通分します。76−1033=7766−2066=5766=1922\frac{7}{6} - \frac{10}{33} = \frac{77}{66} - \frac{20}{66} = \frac{57}{66} = \frac{19}{22}67−3310=6677−6620=6657=22191912÷(94−)=1922\frac{19}{12} \div (\frac{9}{4} - \boxed{}) = \frac{19}{22}1219÷(49−)=221994−=1912÷1922\frac{9}{4} - \boxed{} = \frac{19}{12} \div \frac{19}{22}49−=1219÷221994−=1912×2219\frac{9}{4} - \boxed{} = \frac{19}{12} \times \frac{22}{19}49−=1219×192294−=2212=116\frac{9}{4} - \boxed{} = \frac{22}{12} = \frac{11}{6}49−=1222=611=94−116\boxed{} = \frac{9}{4} - \frac{11}{6}=49−611=2712−2212\boxed{} = \frac{27}{12} - \frac{22}{12}=1227−1222=512\boxed{} = \frac{5}{12}=1253. 最終的な答え512\frac{5}{12}125