Упр.845 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б) 3,25 : 3*1/5 + 6,75*(47/60 — 2*17/45 + 1,65);
в) 12:7*1/2 + 7,5:12 + 1/4:0,4*(5,1-3,86);
г) 12:1*1/2+13,2:11+(0,7:1*3/4)*(0,276:0,23).

Подробный ответ

а)
$$\left(1{,}6-2\frac{1}{6}+\frac{41}{90}\right)\cdot 3\frac{3}{5}-0{,}25:1{,}25$$
$$=\left(1\frac{3}{5}-2\frac{1}{6}+\frac{41}{90}\right)\cdot 3\frac{3}{5}-0{,}2$$
$$=\left(\frac{144}{90}-\frac{195}{90}+\frac{41}{90}\right)\cdot \frac{18}{5}-0{,}2$$
$$=-\frac{10}{90}\cdot \frac{18}{5}-0{,}2=-\frac{1}{9}\cdot \frac{18}{5}-0{,}2=-\frac{2}{5}-0{,}2=-0{,}6.$$
б)
$$3{,}25:3\frac{1}{5}+6{,}75\left(\frac{47}{60}-2\frac{17}{45}+1{,}65\right)$$
$$=3\frac{25}{100}:3\frac{1}{5}+6{,}75\left(\frac{47}{60}-2\frac{17}{45}+1\frac{65}{100}\right)$$
$$=3\frac{1}{4}:3\frac{1}{5}+6{,}75\left(\frac{47}{60}-2\frac{17}{45}+1\frac{13}{20}\right)$$
$$=\frac{13}{4}:\frac{16}{5}+6{,}75\left(\frac{141}{180}-\frac{68}{180}+\frac{117}{180}\right)$$
$$=\frac{13}{4}\cdot \frac{5}{16}+6{,}75\left(\frac{141-68+117}{180}\right)$$
$$=\frac{65}{64}+6{,}75\left(\frac{190}{180}\right)=\frac{65}{64}+6{,}75\cdot \frac{19}{18}$$
$$=\frac{65}{64}+\frac{27}{4}\cdot \frac{1}{18}=\frac{65}{64}+\frac{3}{8}=\frac{65}{64}+\frac{24}{64}=\frac{89}{64}=1\frac{25}{64}.$$
в)
$$12:7\frac{1}{2}+7{,}5:12+\frac{1}{4}:0{,}4\cdot (5{,}1-3{,}86)$$
$$=12:\frac{15}{2}+\frac{15}{2}:12+\frac{1}{4}:\frac{2}{5}\cdot 1{,}24$$
$$=12\cdot \frac{2}{15}+\frac{15}{2}\cdot \frac{1}{12}+\frac{1}{4}\cdot \frac{5}{2}\cdot \frac{124}{100}$$
$$=\frac{8}{5}+\frac{5}{8}+\frac{31}{40}=\frac{64}{40}+\frac{25}{40}+\frac{31}{40}=\frac{120}{40}=3.$$
г)
$$12:1\frac{1}{2}+13{,}2:11+\left(0{,}7:1\frac{3}{4}\right)\cdot (0{,}276:0{,}23)$$
$$=12:\frac{3}{2}+\frac{132}{10}\cdot \frac{1}{11}+\left(\frac{7}{10}:\frac{7}{4}\right)\cdot 1{,}2$$
$$=12\cdot \frac{2}{3}+\frac{12}{10}+\left(\frac{7}{10}\cdot \frac{4}{7}\right)\cdot 1{,}2$$
$$=8+1{,}2+0{,}4\cdot 1{,}2=8+1{,}2+0{,}48=9{,}68.$$