Решебник по математике 6 класс Мерзляк ФГОС Задание 1115

 

\[\boxed{\mathbf{1115\ (1106)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1) - 4 \cdot (n - k) = 4 \cdot (k - n)\]

\[k - n = - 7 \Longrightarrow 4 \cdot (k - n) =\]

\[= 4 \cdot ( - 7) = - 28.\]

\[2)\ 4m - (m + 3n) = 4m - m -\]

\[- 3n = 3m - 3n = 3 \cdot (m - n)\]

\[m - n = - 0,8 \Longrightarrow 3 \cdot (m - n) =\]

\[= 3 \cdot ( - 0,8) = - 2,4.\]

\[3) - 3a - (8b - 15a) = - 3a -\]

\[- 8b + 15a = 12a - 8b =\]

\[= 4 \cdot (3a - 2b)\]

\[3a - 2b = - 0,25 \Longrightarrow 4 \cdot\]

\[\cdot (3a - 2b) = 4 \cdot ( - 0,25) = - 1.\]

\[4)\ 6 \cdot (2x - 3y) - 2 \cdot (x + y) =\]

\[= 12x - 18y - 2x - 2y =\]

\[= 10x - 20y = - 10 \cdot (2y - x)\]

\[2y - x = 17,8 \Longrightarrow - 10 \cdot\]

\[\cdot (2y - x) = - 10 \cdot 17,8 = - 178.\]

\[5)\ 7a(3b + 4c) -\]

\[- 3a\left( b + \frac{1}{3}c \right) = 21ab +\]

\[+ 28ac - 3ab - ac =\]

\[= 18ab + 27ac = 9a(3c + 2b)\]

\[3c + 2b = - 1,6;\ \ a = - 3\frac{1}{3} \Longrightarrow\]

\[9a(3c + 2b) = 9 \cdot (3c + 2b) =\]

\[= 9 \cdot \left( - 3\frac{1}{3} \right) \cdot ( - 1,6) =\]

\[= - 14,4 \cdot \left( - \frac{10}{3} \right) = 48\]