Упр.1 Вариант 3 Дидактические материалы ГДЗ Мерзляк Полонский 7 класс (Алгебра)
- Найдите значение выражения:
1) $$4\frac{1}{7}\cdot14-2\frac{1}{4}\cdot3\frac{1}{6}-1\frac{1}{9}\cdot3\frac{3}{8}$$;
2) $$1\frac{31}{32}\cdot3\frac{1}{5}-\left(8\frac{5}{9}\cdot\frac{6}{35}+2\frac{2}{15}\right)\cdot\frac{5}{12}$$;
3) $$\left(4\frac{5}{12}-3\frac{13}{24}\right):1\frac{3}{4}+\frac{5}{6}:\frac{5}{7}$$;
4) $$\left(2{,}06:\frac{1}{60}-14{,}84:\frac{7}{60}\right)\cdot\frac{1}{6}-0{,}084\cdot\frac{1}{12}$$;
5) $$\left(-16{,}2:32{,}4-21{,}2:(-10{,}6)\right)\cdot(-2{,}8)$$;
6) $$\left(-2{,}3-3{,}91:(-2{,}3)\right):(-0{,}01)\cdot(-0{,}7)$$;
7) $$\left(-\frac{11}{15}-\frac{7}{20}\right):\left(-3\frac{1}{4}\right)$$;
8) $$\left(-\frac{11}{18}+\frac{29}{45}\right):\left(\frac{19}{27}-\frac{35}{54}\right)$$;
9) $$-4\frac{1}{7}+2\frac{1}{4}\cdot\left(-11\frac{2}{9}-(-5{,}4):\frac{9}{35}\right)$$.
1) $$4\frac{1}{7}\cdot 14-2\frac{1}{4}\cdot 3\frac{1}{6}-1\frac{1}{9}\cdot 3\frac{3}{8}$$
$$=\frac{29}{7}\cdot 14-\frac{9}{4}\cdot \frac{19}{6}-\frac{10}{9}\cdot \frac{27}{8}$$
$$=58-\frac{57}{8}-\frac{15}{4}=58-7\frac{1}{8}-3\frac{3}{4}=47\frac{1}{8}.$$
2) $$1\frac{31}{32}\cdot 3\frac{1}{5}-\left(8\frac{5}{9}\cdot \frac{6}{35}+2\frac{2}{15}\right)\cdot \frac{5}{12}$$
$$=\frac{63}{32}\cdot \frac{16}{5}-\left(\frac{77}{9}\cdot \frac{6}{35}+\frac{32}{15}\right)\cdot \frac{5}{12}$$
$$=\frac{63}{10}-\left(\frac{22}{15}+\frac{32}{15}\right)\cdot \frac{5}{12}$$
$$=\frac{63}{10}-\frac{54}{15}\cdot \frac{5}{12}=6\frac{3}{10}-1\frac{1}{2}=4{,}8.$$
3) $$\left(4\frac{5}{12}-3\frac{13}{24}\right):1\frac{3}{4}+\frac{5}{6}:\frac{5}{7}$$
$$=\left(4\frac{10}{24}-3\frac{13}{24}\right):\frac{7}{4}+\frac{5}{6}:\frac{5}{7}$$
$$=\frac{21}{24}:\frac{7}{4}+\frac{5}{6}\cdot \frac{7}{5}=\frac{7}{8}\cdot \frac{4}{7}+\frac{7}{6}$$
$$=\frac{1}{2}+\frac{7}{6}=1\frac{2}{3}.$$
4) $$\left(2{,}06:\frac{1}{60}-14{,}84:\frac{7}{60}\right)\cdot \frac{1}{6}-0{,}084\cdot \frac{1}{12}$$
$$=\left(2{,}06\cdot 60-14{,}84\cdot \frac{60}{7}\right)\cdot \frac{1}{6}-0{,}007$$
$$=\left(123{,}6-127{,}2\right)\cdot \frac{1}{6}-0{,}007$$
$$=-3{,}6\cdot \frac{1}{6}-0{,}007=-0{,}6-0{,}007=-0{,}607.$$
5) $$\left(-16{,}2:32{,}4-21{,}2:(-10{,}6)\right)\cdot (-2{,}8)$$
$$=\left(-\frac{1}{2}+2\right)\cdot \left(-\frac{28}{10}\right)=\frac{3}{2}\cdot \left(-\frac{14}{5}\right)=-\frac{42}{10}=-4{,}2.$$
6) $$\left(-2{,}3-3{,}91:(-2{,}3)\right):(-0{,}01)\cdot (-0{,}7)$$
$$=\left(-2{,}3+1{,}7\right):(-0{,}01)\cdot (-0{,}7)$$
$$=-0{,}6:(-0{,}01)\cdot (-0{,}7)=60\cdot (-0{,}7)=-42.$$
7) $$\left(-\frac{11}{15}-\frac{7}{20}\right):\left(-3\frac{1}{4}\right)$$
$$=\left(-\frac{44}{60}-\frac{21}{60}\right):\left(-\frac{13}{4}\right)=-\frac{65}{60}:\left(-\frac{13}{4}\right)$$
$$=-\frac{13}{12}\cdot \left(-\frac{4}{13}\right)=\frac{1}{3}.$$
8) $$\left(-\frac{11}{18}+\frac{29}{45}\right):\left(\frac{19}{27}-\frac{35}{54}\right)$$
$$=\left(-\frac{55}{90}+\frac{58}{90}\right):\left(\frac{38}{54}-\frac{35}{54}\right)$$
$$=\frac{3}{90}:\frac{3}{54}=\frac{1}{30}:\frac{1}{18}=\frac{18}{30}=\frac{3}{5}=0{,}6.$$
9) $$-4\frac{1}{7}+2\frac{1}{4}\cdot \left(-11\frac{2}{9}-(-5{,}4):\frac{9}{35}\right)$$
$$=-4\frac{1}{7}+\frac{9}{4}\cdot \left(-11\frac{2}{9}+\frac{54}{10}\cdot \frac{35}{9}\right)$$
$$=-4\frac{1}{7}+\frac{9}{4}\cdot \left(-11\frac{2}{9}+21\right)$$
$$=-4\frac{1}{7}+\frac{9}{4}\cdot 9\frac{7}{9}=-4\frac{1}{7}+22=17\frac{6}{7}.$$
Ответ: 1) $$47\frac{1}{8}$$; 2) $$4{,}8$$; 3) $$1\frac{2}{3}$$; 4) $$-0{,}607$$; 5) $$-4{,}2$$; 6) $$-42$$; 7) $$\frac{1}{3}$$; 8) $$\frac{3}{5}$$; 9) $$17\frac{6}{7}$$.
