Страница 401 номер 1243, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\frac{\left( 28\ :1\frac{3}{4} + 7\frac{1}{3}\ :22 + 1\frac{2}{3} \cdot 9\frac{3}{4} + 14\ :1\frac{1}{2} \right) \cdot 3\frac{1}{7}}{10\frac{1}{2} - 9\frac{3}{4}} =\)
\(= \frac{\left( \frac{28 \cdot 4}{7} + \frac{22 \cdot 1}{3 \cdot 22} + \frac{5 \cdot 39}{3 \cdot 4} + \frac{14 \cdot 2}{3} \right) \cdot \frac{22}{7}}{\frac{42}{4} - \frac{39}{4}} = \frac{\left( 16 + \frac{1}{3} + \frac{65}{4} + \frac{28}{3} \right) \cdot \frac{22}{7}}{\frac{3}{4}} =\)
\(= \left( \dfrac{192}{12} + \dfrac{4}{12} + \dfrac{195}{12} + \dfrac{112}{12} \right) \cdot \dfrac{22}{7} \cdot \dfrac{4}{3} = \dfrac{503}{12} \cdot \dfrac{22}{7} \cdot \dfrac{4}{3} = \dfrac{11\ 066}{63};\)

2) \(\left( \frac{1}{2} - 0{,}375 \right)\ :0{,}125 + \left( \frac{5}{6} - \frac{7}{12} \right)\ :(0{,}358 - 0{,}108) =\)
\(= (0{,}5 - 0{,}375)\ :0{,}125 + \dfrac{10 - 7}{12}\ :0{,}25 = 0{,}125\ :0{,}125 + \dfrac{1}{4} \cdot 4 =\)
\(= 1 + 1 = 2.\)