Упр.1243 ГДЗ Алимов 10-11 класс (Алгебра)

1)

$$\frac{\left(28:1\frac{3}{4}+7\frac{1}{3}:22+1\frac{2}{3}\cdot 9\frac{3}{4}+14:1\frac{1}{2}\right)\cdot 3\frac{1}{7}}{10\frac{1}{2}-9\frac{3}{4}}$$
$$28:1\frac{3}{4}=28:\frac{7}{4}=16,\quad 7\frac{1}{3}:22=\frac{22}{3}:22=\frac{1}{3}$$
$$1\frac{2}{3}\cdot 9\frac{3}{4}=\frac{5}{3}\cdot\frac{39}{4}=\frac{65}{4},\quad 14:1\frac{1}{2}=14:\frac{3}{2}=\frac{28}{3}$$
$$16+\frac{1}{3}+\frac{65}{4}+\frac{28}{3} =16+\frac{29}{3}+\frac{65}{4} =16+\frac{116+195}{12} =16+\frac{311}{12} =\frac{503}{12}$$
$$\frac{503}{12}\cdot 3\frac{1}{7}:\left(10\frac{1}{2}-9\frac{3}{4}\right) = \frac{503}{12}\cdot\frac{22}{7}:\left(\frac{21}{2}-\frac{39}{4}\right)$$
$$\frac{21}{2}-\frac{39}{4}=\frac{42-39}{4}=\frac{3}{4}$$
$$\frac{503}{12}\cdot\frac{22}{7}:\frac{3}{4} = \frac{503}{12}\cdot\frac{22}{7}\cdot\frac{4}{3} = \frac{11066}{63} = 175\frac{41}{63}$$

Ответ: $$175\frac{41}{63}$$

2)

$$\left(\frac{1}{2}-0{,}375\right):0{,}125+\left(\frac{5}{6}-\frac{7}{12}\right):(0{,}358-0{,}108)$$
$$\frac{1}{2}-0{,}375=0{,}5-0{,}375=0{,}125,\quad 0{,}125:0{,}125=1$$
$$\frac{5}{6}-\frac{7}{12}=\frac{10}{12}-\frac{7}{12}=\frac{3}{12}=\frac{1}{4},\quad 0{,}358-0{,}108=0{,}25$$
$$\frac{1}{4}:0{,}25=\frac{1}{4}:\frac{1}{4}=1$$
$$1+1=2$$

Ответ: $$2$$