Решебник по математике 6 класс Мерзляк ФГОС Задание 1355

 

\[\boxed{\mathbf{1355}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ при\ m = - 0,3:\]

\[4 \cdot (2 - 3m) - (6 - m) -\]

\[- 2 \cdot (3m + 4) = 8 - 12m -\]

\[- 6 + m - 6m - 8 =\]

\[= - 17m - 6 = - 17 \cdot ( - 0,3) -\]

\[- 6 = 5,1 - 6 = - 0,9.\ \]

\[2)\ при\ n = 0,21:\]

\[- 0,5 \cdot (1 - 3n) +\]

\[+ 4 \cdot (0,2n - 0,1) -\]

\[- (0,1 - 0,7n) =\]

\[= - 0,5 + 1,5n + 0,8n - \ 0,4 -\]

\[- 0,1 + 0,7n = 2,3n + 0,7n - 1 =\]

\[= 3n - 1 = 3 \cdot 0,21 - 1 =\]

\[= 0,63 - 1 = - 0,37.\ \]

\[3)\ при\ x = 1\frac{2}{7} = \frac{9}{7},\ y = - \frac{5}{6}:\]

\[(xy - 2x + 5y) \cdot 3 +\]

\[+ 2 \cdot (2xy + 3x - 6y) =\]

\[= 3xy - 6x + 15y + 4xy + 6x -\]

\[- 12y = 7xy + 3y =\]

\[= 7 \cdot \frac{9}{7} \cdot \left( - \frac{5}{6} \right) + 3 \cdot \left( - \frac{5}{6} \right) =\]

\[= - 3 \cdot \frac{5}{2} - \frac{5}{2} = - 3 \cdot 2,5 - 2,5 =\]

\[= - 7,5 - 2,5 = - 10.\ \]

\[4)\ при\ m = - 1,\ k = - 3,5,\ n = 4:\]

\[(3m - 4n)k - m(6n + 5k) -\]

\[- (nk - 6mn) =\]

\[= 3mk - 4nk - 6mn - 5mk -\]

\[- nk + 6mn = - 2mk - 5nk =\]

\[= - 2 \cdot ( - 1) \cdot ( - 3,5) -\]

\[- 5 \cdot 4 \cdot ( - 3,5) = - 7 + 20 \cdot 3,5 =\]

\[= - 7 + 70 = 63.\ \]

\[5)\ при\ a = 1,5;b = - 2,6:\]

\[2a(3 - b) - 3b(a - 2) -\]

\[- 5 \cdot (ab + a + b) =\]

\[= 6a - 2ab - 3ab + 6b - 5ab -\]

\[- 5a - 5b = a - 5ab -\]

\[- 5ab + b =\]

\[= a - 10ab + b = 1,5 -\]

\[- 10 \cdot 1,5 \cdot ( - 2,6) + ( - 2,6) =\]

\[= 1,5 + 15 \cdot 2,6 - 2,6 =\]

\[{= 1,5 + 39 - 2,6 = 1,5 + }{+ 36,4 = 37,9.\ }\]

\[6)\ при\ m = 10,\ n = \frac{5}{18}:\]

\[- \frac{5}{8} \cdot (5,6m - 1,6n) -\]

\[- 7,2 \cdot \left( - \frac{4}{9}m + 1\frac{7}{18}n \right) =\]

\[= - 5 \cdot 0,7m + 5 \cdot 0,2n +\]

\[+ 0,8 \cdot 4m - 7,2 \cdot \frac{25}{18}n =\]

\[= - 3,5m + n + 3,2m -\]

\[- 0,4 \cdot 25n = - 0,3m + n -\]

\[- 10n = - 0,3m - 9n =\]

\[= - 0,3 \cdot 10 - 9 \cdot \frac{5}{18} = - 3 -\]

\[- \frac{5}{2} = - 3 - 2,5 = - 5,5.\ \]

\[7)\ при\ x = - 1\frac{1}{3} = - \frac{4}{3},\ y = 1,2:\]

\[- \frac{3}{7} \cdot \left( 2,1x + 4\frac{2}{3}y \right) +\]

\[+ 2,2 \cdot \left( - \frac{3}{11}x - \frac{5}{22}y \right) =\]

\[= - 3 \cdot 0,3x - \frac{3}{7} \cdot \frac{14}{3}y -\]

\[- 0,2 \cdot 3x - 0,1 \cdot 5y =\]

\[= - 0,9x - 2y - 0,6x - 0,5y =\]

\[= - 1,5x - 2,5y =\]

\[= - 1,5 \cdot \left( - \frac{4}{3} \right) - 2,5 \cdot 1,2 =\]

\[= 0,5 \cdot 4 - 3 = 2 - 3 = - 1.\ \]

\[8)\ при\ a = 5,5;b = 2\frac{8}{11} = \frac{30}{11}:\]

\[\frac{7}{23} \cdot \left( 3\frac{2}{7}a - 2\frac{4}{21}b \right) -\]

\[- \frac{9}{16} \cdot \left( 5\frac{1}{3}a - \frac{8}{15}b \right) =\]

\[= \frac{7}{23} \cdot \frac{23}{7}a - \frac{7}{23} \cdot \frac{46}{21}b -\]

\[- \frac{9}{16} \cdot \frac{16}{3}a + \frac{9}{16} \cdot \frac{8}{15}b =\]

\[= a - \frac{2}{3}b - 3a + \frac{3}{2} \cdot \frac{1}{5}b =\]

\[= - 2a - \frac{2}{3}b + \frac{3}{10}b =\]

\[= - 2a - \frac{20}{30}b + \frac{9}{30}b = - 2a -\]

\[- \frac{11}{30}b = - 2 \cdot 5,5 - \frac{11}{30} \cdot \frac{30}{11} =\]

\[= - 11 - 1 = - 12.\text{\ \ }\]