Упр.939 ГДЗ Никольский Потапов 7 класс (Алгебра)

1. 7. а) $$\frac{(9{,}126:0{,}65+0{,}46)\cdot7.18+1{,}45\cdot28{,}2}{3{,}45^2-0{,}55^2}$$; б) $$\frac{3{,}05^2-2{,}55^2}{0{,}35\cdot388-28{,}8\cdot(20{,}56-14{,}501:0{,}85)}$$; в) $$\frac{(3\cdot\frac{9}{20}+1\cdot\frac{1}{6}):27{,}7+5\cdot\frac{1}{7}\cdot3{,}85-14\cdot\frac{3}{20}}{(1{,}75:\frac{2}{3}-1{,}75:1\cdot\frac{1}{8}):\frac{7}{12}}$$.

а)

$$\frac{(9{,}126:0{,}65+0{,}46)\cdot 7{,}18+1{,}45\cdot 28{,}2}{3{,}45^2-0{,}55^2} = \frac{(14{,}04+0{,}46)\cdot 7{,}18+1{,}45\cdot 28{,}2}{(3{,}45-0{,}55)(3{,}45+0{,}55)}$$

$$= \frac{14{,}5\cdot 7{,}18+1{,}45\cdot 28{,}2}{2{,}9\cdot 4} = \frac{14{,}5\cdot 7{,}18+14{,}5\cdot 2{,}82}{2{,}9\cdot 4}$$

$$= \frac{14{,}5\cdot (7{,}18+2{,}82)}{2{,}9\cdot 4} = \frac{14{,}5\cdot 10}{11{,}6} = \frac{145}{11{,}6} = 12{,}5$$

б)

$$\frac{3{,}05^2-2{,}55^2}{0{,}35\cdot 388-28{,}8\cdot (20{,}56-14{,}501:0{,}85)} = \frac{(3{,}05-2{,}55)(3{,}05+2{,}55)}{0{,}35\cdot 388-28{,}8\cdot (20{,}56-17{,}06)}$$

$$= \frac{0{,}5\cdot 5{,}6}{0{,}35\cdot 388-28{,}8\cdot 3{,}5} = \frac{0{,}5\cdot 5{,}6}{3{,}5\cdot 38{,}8-3{,}5\cdot 28{,}8}$$

$$= \frac{0{,}5\cdot 5{,}6}{3{,}5\cdot (38{,}8-28{,}8)} = \frac{2{,}8}{3{,}5\cdot 10} = \frac{2{,}8}{35} = 0{,}08$$

в)

$$\frac{(3\cdot \frac{9}{20}+1\cdot \frac{1}{6}):27{,}7+5\cdot \frac{1}{7}\cdot 3{,}85-14\cdot \frac{3}{20}} {(1{,}75:\frac{2}{3}-1{,}75:1\frac{1}{8}):\frac{7}{12}}$$

$$= \frac{\left(\frac{27}{20}+\frac{1}{6}\right):\frac{277}{10}+\frac{5}{7}\cdot \frac{77}{20}-\frac{42}{20}} {\left(1{,}75\cdot \frac{3}{2}-1{,}75\cdot \frac{8}{9}\right):\frac{7}{12}}$$

$$= \frac{\frac{49}{60}\cdot \frac{10}{277}+\frac{11}{4}-\frac{21}{10}} {1{,}75\cdot \left(\frac{3}{2}-\frac{8}{9}\right)\cdot \frac{12}{7}} = \frac{\frac{49}{1662}+\frac{55}{20}-\frac{21}{10}} {1{,}75\cdot \frac{11}{18}\cdot \frac{12}{7}}$$

$$= \frac{\frac{49}{1662}+\frac{13}{20}} {\frac{11}{6}} = \frac{349}{110} = 3\frac{19}{110}$$

Ответ

а) $$12{,}5$$; б) $$0{,}08$$; в) $$3\frac{19}{110}$$.