\(x:4+1125=2285\\ x:4=2285-1125\\ x:4=1160\\ x=1160\times4\\ x=4640\)

\(\dfrac{1}{2},\dfrac{3}{4},\dfrac{3}{8},\dfrac{5}{4}\) MSC : 8

\(\dfrac{1}{2}=\dfrac{4}{8}\)

\(\dfrac{3}{4}=\dfrac{6}{8}\)

\(\dfrac{3}{8}\)

\(\dfrac{5}{4}=\dfrac{10}{8}\)

Vậy số bé nhất là \(\dfrac{3}{8}\)

\(\left(x-\dfrac{1}{2}\right)+75\%=\dfrac{9}{10}\\ \left(x-\dfrac{1}{2}\right)+\dfrac{3}{4}=\dfrac{9}{10}\\ \left(x-\dfrac{1}{2}\right)=\dfrac{9}{10}-\dfrac{3}{4}\\ \left(x-\dfrac{1}{2}\right)=\dfrac{3}{20}\\ x=\dfrac{3}{20}+\dfrac{1}{2}\\ x=\dfrac{13}{20}\)

\(\dfrac{1}{2},\dfrac{2}{3},\dfrac{5}{6},\dfrac{7}{12}\)     MSC : 12

\(\dfrac{1}{2}=\dfrac{6}{12}\\ \dfrac{2}{3}=\dfrac{8}{12}\\ \dfrac{5}{6}=\dfrac{10}{12}\\ \dfrac{7}{12}\)

Vậy số lớn nhất trong các phân số trên là \(\dfrac{5}{6}\)

Số học sinh nữ chiếm số phần trăm học sinh cả lớp là

\(\dfrac{25}{\left(15+25\right)}\times100=62,5\%\)

Bài 3 :

a. \(\dfrac{3}{7}+\dfrac{2}{3}+\dfrac{1}{6}=\dfrac{18}{42}+\dfrac{28}{42}+\dfrac{7}{42}=\dfrac{53}{42}\)

b. \(\dfrac{4}{5}-\dfrac{1}{3}-\dfrac{4}{9}=\dfrac{36}{45}-\dfrac{15}{45}-\dfrac{20}{45}=\dfrac{1}{45}\)

c. \(\dfrac{3}{8}\times\dfrac{3}{2}:6=\dfrac{9}{16}:6=\dfrac{9}{16}\times\dfrac{1}{6}=\dfrac{3}{32}\)

d. \(\dfrac{6}{9}:\dfrac{2}{5}\times\dfrac{1}{3}=\dfrac{6}{9}\times\dfrac{5}{2}\times\dfrac{1}{3}=\dfrac{5}{9}\)

\(-\dfrac{7}{12}.1\dfrac{3}{8}-4\dfrac{5}{8}.\dfrac{7}{12}+\dfrac{1}{2}\\ =-\dfrac{7}{12}.\dfrac{3}{8}-\dfrac{20}{8}.\dfrac{7}{12}+\dfrac{1}{2}\\ =-\dfrac{21}{96}-\dfrac{140}{96}+\dfrac{1}{2}\\ =-\dfrac{21}{96}-\dfrac{140}{96}+\dfrac{48}{96}\\ =-\dfrac{113}{96}\)

a. \(n_{H_2}=\dfrac{6.72}{22,4}=0,3\left(mol\right)\)

PTHH : 2K + 2H₂O -> 2KOH + H₂

           0,6                     0,6       0,3

\(m_K=0,6.39=23,4\left(g\right)\)

\(m_{hh}=23,4+14,4=37,8\left(g\right)\)

b. \(C_M=\dfrac{0.6}{2}=0,3\left(M\right)\)

c. \(n_{FeO}=\dfrac{14.4}{72}=0,2\left(mol\right)\)

\(n_{HCl}=\dfrac{182,5.10\%}{36,5}=0,5\left(mol\right)\)

PTHH : FeO + 2HCl -> FeCl₂ + H₂

                0,2                     0,2

Ta thấy : \(\dfrac{0.2}{1}< \dfrac{0.5}{2}\) => FeO đủ , HCl dư

\(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

\(C\%=\dfrac{25,4}{14,4+182,5}.100\%=12,8\%\)

\(n_{H_2}=\dfrac{2.24}{22,4}=0,1\left(mol\right)\)

PTHH : Ca + 2H₂O -> Ca(OH)₂ + H₂

           0,1                                     0,1  

PTHH : CaO + H₂O -> Ca(OH)₂

\(m_{Ca}=0,1.40=4\left(g\right)\)

\(m_{CaO}=16-4=12\left(g\right)\)