Упр.939 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б) (3,05^2 — 2,55^2)/(0,35*388 — 28,8*(20,56-14,501:0,85));
в) ((3*9/20+1*1/6):27,7+5*1/7 * 3,85 — 14*3/20)/((1,75:2/3-1,75:1*1/8):7/12).

Подробный ответ

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)

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

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

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

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