Найдите значение выражений 24y^3 - 3(8y^2-1)(y+6) при y= -2/3 40 m^3 - (5m^2 + m - 2...

$24y^3 - 3(8y^2-1)(y+6)$ при y= - $\frac{2}{3}$
$24y^3 - 3(8y^2-1)(y+6)$ =
= $24y^3 - 3(8y^3 - y + 48 y^{2} - 6 ) =$
= $24y^3 - 24y^3 + 3y - 144 y^{2} + 18 =$
= $- 144 y^{2} + 3y + 18$

y= - $\frac{2}{3}$
$- 144 * (-\frac{2}{3}) ^{2} + 3* (- \frac{2}{3}) + 18 =$
$- 144 * \frac{4}{9} - 2 + 18 = - 64 + 16 = -48$

$40 m^3 - (5m^2 + m - 2 )(8m + 3)$ при m= $\frac{7}{10}$

$40 m^3 - (5m^2 + m - 2 )(8m + 3) =$
= $40 m^3 - (40m^3 + 15 m^{2} + 8m^{2} + 3m - 16m - 6) =$
= $40 m^3 - (40m^3 + 23 m^{2} - 13m - 6) =$
= $40 m^3 - 40m^3 - 23 m^{2} + 13m + 6 =$
= $- 23 m^{2} + 13m + 6$

m= $\frac{7}{10}$ = 0.7
- 23 m^{2} + 13m + 6 =
= $- 23* (0.7)^{2} + 13*(0.7) + 6 =$
= - 23 * 0.49 + 9.1 + 6 =
= -11.27 + 15.1 = 3.83