Упр.864 ГДЗ Муравин 6 класс (Математика)

1)-7/11•1 5/17 :(-0,75);
2) (1,2-2 7/25) :1 2/25+1 4/13;
3) (4,25•16/51-17/18 :8,5)•(3/11)^2;
4) (-12,6•1 1/9+10,8) :6 6/7+2 3/10;
5) (1-7/15+0,4) :(0,3+3/19-15/38);
6) 2:(1 1/5-2,2)+1/10 :(2,8-3/8);
7) (15/98-17/28) :1 40/49+(0,1-0,15)•21 3/7;
8) (5,4•(-3 1/3)+13,8) :1 13/15+3 5/6.

1) $$-\frac{7}{11}\cdot 1\frac{5}{17}:(-0{,}75)=-\frac{7}{11}\cdot \frac{22}{17}:\left(-\frac{3}{4}\right)=\frac{7}{1}\cdot \frac{2}{17}\cdot \frac{4}{3}=\frac{56}{51}=1\frac{5}{51}.$$

2) $$\left(1{,}2-2\frac{7}{25}\right):1\frac{2}{25}+1\frac{4}{13}=\left(1\frac{1}{5}-2\frac{7}{25}\right):\frac{27}{25}+1\frac{4}{13}$$
$$=\left(1\frac{5}{25}-2\frac{7}{25}\right):\frac{27}{25}+1\frac{4}{13}=-1\frac{2}{25}\cdot \frac{25}{27}+1\frac{4}{13}$$
$$=-\frac{27}{25}\cdot \frac{25}{27}+1\frac{4}{13}=-1+1\frac{4}{13}=\frac{4}{13}.$$

3) $$\left(4{,}25\cdot \frac{16}{51}-\frac{17}{18}:8{,}5\right)\cdot \left(\frac{3}{11}\right)^2$$
$$=\left(4\frac{1}{4}\cdot \frac{16}{51}-\frac{17}{18}:8\frac{1}{2}\right)\cdot \frac{9}{121}$$
$$=\left(\frac{17}{4}\cdot \frac{16}{51}-\frac{17}{18}:\frac{17}{2}\right)\cdot \frac{9}{121}$$
$$=\left(\frac{4}{3}-\frac{1}{9}\right)\cdot \frac{9}{121}=\frac{11}{9}\cdot \frac{9}{121}=\frac{1}{11}.$$

4) $$\left(-12{,}6\cdot 1\frac{1}{9}+10{,}8\right):6\frac{6}{7}+2\frac{3}{10}$$
$$=\left(-\frac{126}{10}\cdot \frac{10}{9}+10{,}8\right):\frac{48}{7}+2\frac{3}{10}$$
$$=(-14+10{,}8)\cdot \frac{7}{48}+2\frac{3}{10}=-3{,}2\cdot \frac{7}{48}+2\frac{3}{10}$$
$$=-\frac{16}{5}\cdot \frac{7}{48}+2\frac{3}{10}=-\frac{7}{15}+2\frac{3}{10}$$
$$=-\frac{14}{30}+\frac{69}{30}=\frac{55}{30}=1\frac{5}{6}.$$

5) $$\left(1-\frac{7}{15}+0{,}4\right):\left(0{,}3+\frac{3}{19}-\frac{15}{38}\right)$$
$$=\left(\frac{8}{15}+\frac{6}{15}\right):\left(\frac{3}{10}+\frac{6}{38}-\frac{15}{38}\right)$$
$$=\frac{14}{15}:\left(\frac{57}{190}-\frac{45}{190}\right)=\frac{14}{15}:\frac{12}{190}$$
$$=\frac{14}{15}\cdot \frac{190}{12}=\frac{7}{3}\cdot \frac{19}{6}=\frac{133}{18}=7\frac{7}{18}.$$

6) $$2:\left(1\frac{1}{5}-2{,}2\right)+\frac{1}{10}:\left(2{,}8-\frac{3}{8}\right)$$
$$=2:(1{,}2-2{,}2)+\frac{1}{10}:\left(2\frac{4}{5}-\frac{3}{8}\right)$$
$$=2:(-1)+\frac{1}{10}:\left(\frac{112}{40}-\frac{15}{40}\right)=-2+\frac{1}{10}:\frac{97}{40}$$
$$=-2+\frac{1}{10}\cdot \frac{40}{97}=-2+\frac{4}{97}=-1\frac{93}{97}.$$

7) $$\left(\frac{15}{98}-\frac{17}{28}\right):1\frac{40}{49}+\left(0{,}1-0{,}15\right)\cdot 21\frac{3}{7}$$
$$=\left(\frac{30}{196}-\frac{119}{196}\right):\frac{89}{49}+(-0{,}05)\cdot \frac{150}{7}$$
$$=-\frac{89}{196}\cdot \frac{49}{89}-\frac{1}{20}\cdot \frac{150}{7}=-\frac{1}{4}-\frac{15}{14}$$
$$=-\frac{7}{28}-\frac{30}{28}=-\frac{37}{28}=-1\frac{9}{28}.$$

8) $$\left(5{,}4\cdot \left(-3\frac{1}{3}\right)+13{,}8\right):1\frac{13}{15}+3\frac{5}{6}$$
$$=\left(5\frac{2}{5}\cdot \left(-\frac{10}{3}\right)+13{,}8\right):\frac{28}{15}+3\frac{5}{6}$$
$$=\left(\frac{27}{5}\cdot \left(-\frac{10}{3}\right)+13{,}8\right)\cdot \frac{15}{28}+3\frac{5}{6}$$
$$=(-18+13{,}8)\cdot \frac{15}{28}+3\frac{5}{6}=-4{,}2\cdot \frac{15}{28}+3\frac{5}{6}$$
$$=-\frac{21}{5}\cdot \frac{15}{28}+3\frac{5}{6}=-\frac{9}{4}+3\frac{5}{6}$$
$$=-2\frac{1}{4}+3\frac{5}{6}=1\frac{7}{12}.$$

Ответ

1) $$1\frac{5}{51}$$; 2) $$\frac{4}{13}$$; 3) $$\frac{1}{11}$$; 4) $$1\frac{5}{6}$$; 5) $$7\frac{7}{18}$$; 6) $$-1\frac{93}{97}$$; 7) $$-1\frac{9}{28}$$; 8) $$1\frac{7}{12}$$.